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ORiDV      0  L      TUC      I1UIUCDCITV      R  C      PHI 


PLANE    AND    SOLID 


GEOMETRY 


BY 


EDWARD    RUTLEDGE   ROBBINS,   A.B. 

SENIOR    MATHEMATICAL    MASTER 
THE   WILLIAM    PENN   CHARTER   SCHOOL      > 


NEW  YORK  •:•  CINCINNATI  •:•  CHICAGO 

AMERICAN    BOOK    COMPANY 


COPYRIGHT,  1906,  1907,  BY 
EDWARD  RUTLEDGE  ROBBINS. 

PLANE   AND  80HD   GEOMBTEY. 

w.  P.    3 


FOR    THOSE    WHOSE    PRIVILEGE 
IT   MAY    BE    TO    ACQUIRE    A    KNOWLEDGE    OP 

GEOMETRY 

THIS   VOLUME    HAS    BEEN   WRITTEN 

AND    TO    THE    BOYS    AND    GIRLS    WHO    LEARN   THE    ANCIENT    SCIENCB 

FROM    THESE    PAGES,    AND   WHO    ESTEEM    THE    POWER 

OF    CORRECT    REASONING    THE    MORE 

BECAUSE    OF    THE    LOGIC    OF 

PURE   GEOMETRY 

THIS    VOLUME     IS     DEDICATED 


PREFACE 

THE  motives  actuating  the  author  in  the  preparation  of 
this  text  in  Geometry  have  been  : 


To  present  a  book  that  has  been  written  for  the  pupil. 

The  object  sought  in  the  study  of  Geometry  is  not  solely 
to  train  the  mind  to  accept  only  those  statements  as  truth 
for  which  convincing  reasons  can  be  provided,  but  to  culti- 
vate a  foresight  that  will  appreciate  both  the  purpose  in  mak- 
ing a  statement  and  the  process  of  reasoning  by  which  the 
ultimate  truth  is  established.  Thus,  the  study  of  this  formal 
science  should  develop  in  the  pupil  the  ability  to  pursue 
argument  coherently,  and  to  establish  one  truth  by  the  aid 
of  other  known  truths,  in  logical  order. 

The  more  mature  members  of  a  class  do  not  require  that 
the  reason  for  every  declaration  be  given  in  full  in  the  text  ; 
still,  to  omit  it  altogether,  wrongs  those  pupils  who  do 
not  know  and  cannot  perceive  the  correct  reason.  But  to 
ask  for  the  reason  and  to  print  the  paragraph  reference  meets 
the  requirements  of  the  various  degrees  of  intellectual  capac- 
ity and  maturity  in  every  class.  The  pupil  who  knows  and 
knows  that  he  knows  need  not  consult  the  paragraph  cited  ; 
the  pupil  who  does  not  know  may  learn  for  himself  the 
correct  reason  by  the  reference.  It  is  obvious  that  the 
greater  progress  an  individual  makes  in  assimilating  the  sub- 
ject and  in  entering  into  its  spirit,  the  less  need  there  will 

be  for  the  printed  reference. 

5 


6  PREFACE 

(5)    To  stimulate  the  mental  activity  of  the  pupil. 

To  compel  a  young  student  to  supply  his  own  demonstra- 
tions, in  other  words,  to  think  and  reason  for  himself,  fre- 
quently proves  unprofitable  as  well  as  unpleasant,  and 
engenders  in  the  learner  a  distaste  for  a  study  he  has  the 
right  to  admire  and  to  delight  in.  The  short-sighted  youth 
absorbs  his  Geometry  by  memorizing,  only  to  find  that  his 
memory  has  been  an  enemy,  and  while  he  himself  is  becom- 
ing more  and  more  confused,  his  thoughtful  companion  is 
making  greater  and  greater  progress.  The  earlier  he  dis- 
covers his  error  the  better,  and  the  plan  of  this  text  gives 
him  an  opportunity  to  reestablish  himself  with  his  class. 
It  is  not  calculated  to  produce  accomplished  geometricians 
at  the  completion  of  the  first  book,  but  to  aid  the  learner 
in  his  progress  throughout  the  volume,  wherever  experience 
has  shown  that  he  is  likely  to  require  assistance.  It  is  cal- 
culated, under  good  instruction,  to  develop  a  clear  concep- 
tion of  the  geometric  idea,  and  to  produce  at  the  end  of  the 
course  a  rational  individual  and  a  friend  of  this  particular 
science. 

(#)  To  bring  the  pupil  to  the  theorems  and  their  demonstra- 
tions —  the  real  subject-matter  of  Geometry  —  as  early  in  the 
study  as  possible. 

(c?)  To  explain  rather  than  formally  demonstrate  the  simple 
fundamental  truths. 

(e)  To  apply  each  theorem  in  the  demonstration  of  other 
theorems  as  promptly  as  possible. 

(/)  To  present  a  text  that  will  be  clear,  consistent,  teach- 
able, and  sound. 


PREFACE  T 

The  experienced  teacher  will  observe : 
(a)    The  economy  of  arrangement. 

Many  of  the  smaller  figures  are  placed  at  the  side  of  the 
page  rather  than  at  the  center.  The  individual  numbers  of 
theorems  are  omitted. 

(5)    The  superior  character  of  the  diagrams. 

(<?)    The  omission  of  the  words  "  since  "  and  "for" 

The  advance  statement  is  made  and  the  reason  asked  for 
and  usually  cited.  The  inquiring  mind  fails  to  understand 
the  force  of  preceding  and  following  some  statements  with 
the  same  reason. 

(c?)    Originals   that  are    carefully   classified,  graded,    and 
placed  after  the  natural  subdivisions  of  the  subject-matter. 
(e)    The  independence  of  these  originals. 

Every  exercise  can  be  solved  or  demonstrated  without 
the  use  of  any  other  exercise.  Only  the  truths  in  the 
numbered  paragraphs  are  necessary  in  working  originals. 

(/)  The  setting  of  every  theorem,  corollary,  and  problem  of 
the  text  proper  infullface  type. 

(^)  The  consistent  use  of  such  terms  as  " vertical  angles" 
" vertex- angle"  "adjacent  angles"  "angles  adjoining  a  side" 
and  others. 

(h)  The  full  treatment  of  measurement  and  the  illustrations 
of  the  terms  employed. 

(i)  The  summaries  that  precede  earlier  collections  of  origi- 
nal exercises. 

(/)  The  emphasis  given  to  the  discussion  of  original  con- 
structions- 


8  PREFACE 

As  in  all  subjects  that  are  new  to  a  class,  the  successful 
teacher  will  be  content  with  short  lessons  at  the  beginning, 
and  will  progress  slowly  until  the  class  is  thoroughly  familiar 
with  the  language  and  the  general  method  and  purpose  of 
the  new  science. 

The  author  sincerely  desires  to  extend  his  thanks  to  those 
friends  who,  by  suggestion  and  encouragement,  have  inspired 
him  in  the  preparation  of  these  pages. 

EDWARD  R.   ROBBINS. 

THE  WILLIAM  PENN  CHARTER  SCHOOL, 
PHILADELPHIA. 


CONTENTS 


INTRODUCTION 


PAGE 
11 


Angles 12 

Triangles 14 

Symbols 16 

Axioms 16 

Postulates 17 

BOOK  I.     ANGLES,   LINES, 

RECTILINEAR  FIGURES  .  18 

Preliminary  Theorems  ...  18 
Theorems     and     Demonstra- 
tions        20 

Model  Demonstrations  ...  24 

Quadrilaterals 45 

Polygons 54 

Symmetry 58 

Locus 60 

Summary.  General  Directions 

for  attacking  Originals   .  62 

Original  Exercises    ....  64 

BOOK  II.     THE  CIRCLE     .  79 

Preliminary  Theorems  ...  81 
Theorems     and     Demonstra- 
tions        82 

Summary 92 

Original  Exercises     ....  93 
Kinds  of  Quantities.     Meas- 
urement       97 

Original  Exercises     ....  107 

Constructions 115 

Analysis 127 

Original  Constructions  .     .     .  128 


BOOK    III.      PROPORTION. 
SIMILAR   FIGURES      .     .     140 

Theorems  and  Demonstra- 
tions .......  141 

Original  Exercises  (Numeri- 
cal)   168 

Summary.  Original  Exer- 
cises (Theorems)  ...  172 

Constructions 180 

Original  Constructions  .     .     .     184 


BOOK  IV.     AREAS 


187 


Theorems     and     Demonstra- 
tions        187 

Original     Exercises     (Theo- 
rems)       197 

Formulas 201 

Original  Exercises  (Numeri- 
cal)    204 

Constructions 207 

Original  Constructions  .     .     .  213 


BOOK  V.    REGULAR  POLY- 
GONS.    CIRCLES 


217 


Theorems     and    Demonstra- 
tions       217 

Original      Exercises     (Theo- 
rems)    ........     230 

Constructions 233 

Formulas 236 

Original  Exercises  (Numeri- 
cal)  239 


9 


10 


CONTENTS 


Original  Constructions 
Maxima  and  Minima 
Original  Exercises    . 


PAGE 

244 
245 
249 


BOOK  VI.    LINES,  PLANES, 

AND  ANGLES  IN  SPACE  251 

Preliminary  Theorems       .     .  252 
Theorems     and     Demonstra- 
tions        254 

Original  Exercises    ....  268 

Dihedral  Angles 273 

Original  Exercises     ....  282 

Polyhedral  Angles     ....  284 

Original  Exercises    ....  290 

BOOK  VII.    POLYHEDRONS  291 

Prisms 292 

Preliminary  Theorems  .     .     .  294 
Theorems     and     Demonstra- 
tions        295 

Original  Exercises    ....  304 

Pyramids 307 

Preliminary  Theorems  ...  308 
Theorems     and     Demonstra- 
tions        309 

Regular    and    Similar    Poly- 
hedrons        323 

Formulas 328 

Original  Exercises    ....  329 


BOOK    VIII.      CYLINDERS, 

CONES 337 

Cylinders 337 

Preliminary  Theorems  .     .     .  338 
Theorems     and     Demonstra- 
tions     .     , 339 

Formulas 342 

Original  Exercises     ....  344 

Cones 345 

Preliminary  Theorems  .     .     .  347 
Theorems     and     Demonstra- 
tions        348 

Formulas 353 

Original  Exercises     ....  355 

BOOK  IX.     THE  SPHERE  .  360 

Preliminary  Theorems  .     .     .  362 
Theorems     and    Demonstra- 
tions        363 

Constructions 368 

Spherical  Triangles  ....  370 
Preliminary  Theorems  .     .     .  372 
Theorems     and    Demonstra- 
tions        373 

Original  Exercises    ....  382 
Areas  and  Volumes  ....  386 
Preliminary  Theorems  .     .     .  388 
Theorems     and     Demonstra- 
tions        389 

Original  Exercises     ....  397 

Index  of  Definitions      .    <     ,  405 


PLANE  GEOMETRY 


INTRODUCTION 

1.  Geometry  is  a  science  which  treats  of  the  measurement 
of  magnitudes. 

2.  A  definition  is  a  statement  explaining  the  significance 
of  a  word  or  a  phrase. 

Every  definition  should  be  clear,  simple,  descriptive,  and  correct ;  that 
is,  it  should  contain  the  essential  qualities  or  exclude  all  others,  or  both. 

3.  A  point  is  that  which  has  position  but  not  magnitude. 

4.  A  line  is  that  which  has  length  but  no  other  magnitude. 

5.  A  straight  line  is  a  line  which  is  determined  (fixed  in 
position)  by  any  two  of  its  points.     That  is,  two  lines  that 
coincide  entirely,  if   they  coincide  at  any  two   points,  are 
straight  lines. 

6.  A  rectilinear  figure  is  a  figure  containing  straight  lines 
and  no  others. 

7.  A  surface  is  that  which  has  length  and  breadth  but  no 

other  magnitude. 

8.  A  plane  is  a  surface  in  which  if  any  two  points  are 
taken,  the  straight  line  connecting  them  lies  wholly  in  that 
surface. 

9.  Plane  Geometry  is  a  science  which  treats  of  the  proper- 
ties of  magnitudes  in  a  plane. 

10.   A  solid  is  that  which  has  length,  breadth,  and  thick- 
ness.    A  solid  is  that  which  occupies  space. 

11 


12  PLANE   GEOMETRY 

11.  Boundaries.     The  boundaries  (or  boundary)  of  a  solid 
are  surfaces.     The  boundaries  (or  boundary)  of  a  surface 
are   lines.     The   boundaries   of   a   line   are   points.     These 
boundaries  can  be  no  part  of  the  things  they  limit.     A  sur- 
face is  no  part  of  a  solid  ;  a  line  is  no  part  of  a  surface  ;  a 
point  is  no  part  of  a  line. 

12.  Motion.     If  a  point  moves,  its  path  is  a  line.     Hence, 
if  a  point  moves,  it  generates  (describes  or  traces)  a  line ;  if 
a  line  moves  (except  upon   itself),  it  generates   a  surface  ; 
if  a  surface  moves  (except  upon  itself),  it  generates  a  solid. 

NOTE.     Unless  otherwise  specified  the  word  "  line  "  hereafter  means 
straight  line, 

ANGLES 


ADJACENT     VERTICAL  ANGLES      RIGHT  ANGLES 
ANGLES  PERPENDICULAR 


13.  A  plane  angle  is  the  amount  of  divergence  of  two 
straight  lines   that   meet.      The  lines  are    called    the    sides 
of  the  angle.     The  vertex  of  an  angle  is  the  point  at  which 
the  lines  meet. 

14.  Adjacent   angles  are  two  angles  that  have  the  same 
vertex  and  a  common  side  between  them. 

15.  Vertical    angles    are  two  angles  that  have  the  same 
vertex,  the  sides  of  one  being  prolongations  of  the  sides  of 
the  other. 

16.  If  one  straight  line  meets  another  and  makes  the  ad- 
jacent angles  equal,  the  angles  are  right  angles. 


INTRODUCTION  13 

17.  One  line  is  perpendicular  to  another  if  they  meet  at 
right  angles.     Either  line  is  perpendicular  to  the  other.     The 
point  at  which  the  lines  meet  is  the  foot  of  the  perpendicular. 
Oblique  lines  are  lines  that  meet  but  are  not  perpendicular. 

18.  A  straight  angle  is  an  angle  whose  sides  lie  in  the 
same  straight  line,  but  extend  in  opposite   directions  from 
the  vertex. 


OBTUSE  ANGLE        ACUTE     COMPLEMENTARY  SUPPLEMENTARY  ANGLES 
ANGLE         ANGLES 

19.  An  obtuse  angle  is  an  angle  that  is  greater   than   a 
right  angle.     An  acute  angle  is  an  angle  that  is  less  than 
a    right    angle.     An    oblique    angle   is    any    angle  that   is 
not  a  right  angle. 

20.  Two  angles  are  complementary  if  their  sum  is  equal  to 
one  right  angle.     Two  angles  are  supplementary  if  their  sum 
is  equal  to  two  right  angles.     Thus  the  complement  of  an 
angle  is  the  difference  between  one  right  angle  and  the  given 
angle.    The  supplement  of  an  angle  is  the  difference  between 
two  right  angles  and  the  given  angle. 

21.  A  degree   is  one   ninetieth   of   a   right   angle.     The 
degree  is  the  familiar  unit  used  in  measuring  angles.     It  is 
evident  that  there  are   90°  in  a  right  angle ;    180°  in  two 
right  angles,  or  a  straight  angle ;  360°  in  four  right  angles. 

22.  Notation.   A  point  is  usually  denoted  by  a  capital  letter,  placed 
near  it.     A  line  is  denoted  by  two  capital  letters,  placed  one  at  each  end, 
or  one  at  each  of  two  of  its  points.     Its  length  is  sometimes  represented 
advantageously  by  a  small  letter  written  near  it.     Thus,  the  line  AB ; 
the  line  R  S;  the  line  m. 

A B  5 §_      ™ 


14 


PLANE   GEOMETRY 


An  angle  is  usually  denoted  by  three  capital  letters,  placed  one  at  the 
vertex  and  one  on  each  side.  If  only  one  angle  is  at  a  vertex,  the  capital 
letter  at  the  vertex  is  sufficient  to  designate  the  angle.  Sometimes  it  is 
advantageous  to  name  an  angle  by  a  small  letter  placed  within  the  angle. 
The  word  "  angle  "  is  usually  denoted  by  the  symbol  "  Z "  in  geometrical 
processes. 


X  o  C 

Z  AMX OR  Z  a  AND  Z  BOG, 

Z.XMA  OR  Z M.          NOT  Z  0 


/.x 


It  is  important  that  in  naming  an  angle  by  the  use  of  three  letters,  the 
vertex-letter  should  be  placed  between  the  others.  The  size  of  an  angle 
does  not  depend  upon  the  length  of  the  sides,  but  only  on  the  amount  of 
their  divergence.  Thus,  Z  x  =  Z  P  and  Z  P  is  the  same  as  Z  APR  or 
Z  APS  or  Z  EPS,  etc.  An  angle  is  said  to  be  included  by  its  sides.  An 
angle  is  bisected  by  a  line  drawn  through  the  vertex  and  dividing  the 
angle  into  two  equal  angles. 

TRIANGLES 

23.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines.  These  lines  are  the  sides.  The  vertices  of  a 
triangle  are  the  three  points  at  which  the  sides  intersect. 
The  angles  of  a  triangle  are  the  three  angles  at  the  three 
vertices.  Each  side  of  a  triangle  has  two  angles  adjoining 
it.  The  symbol  for  triangle  is  "A". 


ISOSCELES  A 


EQUILATERAL  A 
EQUIANGULAR  A 


RIGHT  A         OBTUSE  A  ACUTE  A 

SCALENE  & 


The  base  of  a  triangle  is  the  side  on  which  the  figure  appears 
to  stand.  The  vertex  of  a  triangle  is  the  vertex  opposite 
the  base.  The  vertex  -angle  is  the  angle  opposite  the  base. 


INTRODUCTION  16 

24.  Kinds  of  triangles  : 

A  scalene  triangle  is  a  triangle  no  two  sides  of  which  are  equal. 

An  isosceles  triangle  is  a  triangle  two  sides  of  which  are  equal. 

An  equilateral  triangle  is  a  triangle  all  sides  of  which  are  equal. 

A  right  triangle  is  a  triangle  one  angle  of  which  is  a  right  angle. 

An  obtuse  triangle  is  a  triangle  one  angle  of  which  is  an  obtuse  angle. 

An  acute  triangle  is  a  triangle  all  angles  of  which  are  acute  angles. 

An  equiangular  triangle  is  a  triangle  all  angles  of  which  are  equal. 

25.  The  hypotenuse  of  a  right  triangle  is  the  side  op- 
posite the  right  angle.     The  sides  forming  the  right  angle 
are  called  the  legs.     In  an  isosceles  triangle  the  equal  sides 
are  sometimes  called  the  legs,  and  the  other  side,  the  base. 

26.  Homologous  Parts.      If  two  triangles  have  the  three 
angles  of  one  equal  respectively  to  the  three  angles  of  the 
other,  the  pairs  of  equal  angles  are  homologous.     Homologous 
sides  in  two  triangles  are  opposite  the  homologous  angles. 

27.  Homologous  parts  of  equal  figures  are  equal. 

If    the  triangles  DEF  and 
HIJ  are  equal  in  all  respects, 
Z  D  is  homologous  to,  and 
=  Z  H,  hence  EF  is  homolo- 
gous   to,    and  =    IJ.      And      D 
Z  E  is  homologous  to,  and  =  Z  /,  hence,  DF  is  homologous 
to,  and  =  jffj,  and  so  on. 

SUPERPOSITION.    SYMBOLS 

28.  Equality  and  coincidence.      Two   geometrical   figures 
are* equal  if  they  can  be  made  to  coincide  in  all  respects. 
Angles  coincide,  and  are  equal,  if  their  vertices  are  the  same 
point  and  the  sides  of  one  angle  are  identical  with  the  sides 
of  the  other.     Superposition  is  the  process  of  placing  one 
figure  upon  another.     This  method  of  showing  the  equality 
of  two  geometrical  figures  is  employed  only  in  establishing 
fundamental  principles. 


16 


PLANE   GEOMETRY 


29.    Symbols.    The  usual  symbols  and  abbreviations  em< 
ployed  in  geometry  are  the  following : 


+  plus. 

—  minus. 

=  equals,  or  is  (or  are) 
equal  to. 

=3=  is  (or  are)  equivalent  to. 

>  is  (or  are)  greater  than. 

<  is  (or  are)  less  than. 

.*.  hence,  therefore,  conse- 
quently. 

_L  perpendicular. 

Js  perpendiculars. 

Q  circle. 


©  circles. 
Z  angle. 
A  angles. 

rt.  Z  right  angle, 
rt.  A  right  angles. 
A  triangle. 
&  triangles, 
rt.  &  right  triangles. 
||  parallel. 
|| s  parallels. 
O  parallelogram. 
(U  parallelograms. 


ax.  axiom, 

hyp.  hypothesis, 

comp.  complementary. 

supp.  supplementary, 

const,  construction, 

cor.  corollary, 

st.  straight, 

def.  definition, 

alt.  alternate, 

int.  interior, 

ext.  exterior. 


AXIOM,    POSTULATE,    AND    THEOREM 

30.  An  axiom  is  a  truth  assumed  to  be  self-evident.    It  is 
a  truth  which  is  received  and  assented  to  immediately. 

31.  AXIOMS. 

1.  Magnitudes  that  are  equal  to  the  same  thing,  or  to  equals,  are 
equal  to  each  other. 

2 .  If  equals  are  added  to,  or  subtracted  from,  equals,  the  results  are 
equal. 

3.  If  equals  are  multiplied  by,  or  divided  by,  equals,  the  results  are 
equal, 

[Doubles  of  equals  are  equal ;  halves  of  equals  are  equal.] 

4.  The  whole  is  equal  to  the  sum  of  all  of  its  parts. 

5.  The  whole  is  greater  than  any  of  its  parts. 

6.  A  magnitude  may  be  displaced  by  its  equal  in  any  process. 
[Briefly  called  "  substitution."] 

7.  If  equals  are  added  to,  or  subtracted  from,  unequals,  the  results 
are  unequal  in  the  same  sense. 

8.  If  unequals  are  added  to  unequals  in  the  same  sense,  the  results 
are  unequal  in  that  sense. 

9.  If  unequals  are  subtracted  from  equals,  the  results  are  unequal 
in  the  opposite  sense. 


INTRODUCTION  17 

10.  Doubles  or  halves  of  unequals  are  unequal  in  the  same  sense. 

11.  If  the  first  of  three  magnitudes  is  greater  than  the  second,  and 
the  second  is  greater  than  the  third,  the  first  is  greater  than  the  third. 

12.  A  straight  line  is  the  shortest  line  that  can  be  drawn  between 
two  points. 

13.  A  geometrical  figure  may  be  moved  from   one  position  to 
another  without  any  change  in  form  or  magnitude. 

32.  A  postulate  is  something  required  to  be  done,  the  pos- 
sibility of  which  is  admitted  as  evident. 

33.  POSTULATES. 

1.  It  is  possible  to  draw  a  straight  line  from  any  point  to  any 
other  point. 

2.  It  is  possible  to  extend  (prolong  or  produce)  a  straight  line  in- 
definitely, or  to  terminate  it  at  any  point. 

34.  A  geometric  proof  or  demonstration  is  a  logical  course 
of  reasoning  by  which  a  truth  becomes  evident. 

35.  A  theorem  is  a  statement  that  requires  proof. 

In  the  case  of  the  preliminary  theorems  which  follow,  the 
proof  is  very  simple  ;  but  as  these  theorems  are  not  self- 
evident  they  cannot  be  classified  with  the  axioms. 

A  corollary  is  a  truth  immediately  evident,  or  readily  estab- 
lished, from  some  other  truth  or  truths. 


EXERCISE  1.   Draw  an  ZABC.    In  ^ABC  draw  line  BD. 

What  does  Z  ABD  +  Z  DBC  =  ? 

What  does  Z  ABC  -ZABD  =  1 

Ex.  2.  In  a  rt.  Z  ABC  draw  line  BD. 

If  Z  ABD=25°,  how  many  degrees  are  there  in  Z  DBC  ? 

How  many  degrees  are  there  in  the  complement  of  an  angle  of  38°  ? 
How  many  degrees  are  there  in  the  supplement  ? 

Ex.  3.    Draw  a  straight  line  AB  and  take  a  point  X  on  it. 

What  line  does  AX  +  BX  =  ? 

What  line  does  AB  -  BX  =  ? 

Ex.  4.  Draw  a  straight  line  AB  and  prolong  it  to  X  so  that  BX  —  AB. 
Prolong  it  so  that  AX  -  AB. 


BOOK   I 
ANGLES,  LINES,  RECTILINEAR  FIGURES 

PRELIMINARY   THEOREMS 

36.  A  right  angle  is  equal  to  half  a  straight  angle. 
Because  of  the  definition  of  a  right  angle.     (See  16.) 

37.  A  straight  angle  is  equal  to  two  right  angles.     (See  36.) 

38.  Two  straight  lines  can  intersect  in  only  one  point. 
Because  they  would  coincide   entirely  if  they   had  two 

common  points.    (See  5.) 

39.  Only  one  straight  line  can  be  drawn  between  two  points. 
(See  5.) 

40.  A  definite  (limited  or  finite)  straight  line  can  have  only  one 
midpoint. 

Because  the  halves  of  a  line  are  equal. 

41.  All  straight  angles  are  equal. 

Because  they  can  be  made  to  coincide.  (See  28  and  Ax. 
13.) 

42.  All  right  angles  are  equal. 

They  are  halves  of  straight  angles  (36),  and  hence  equal 
(Ax.  3). 

43.  Only  one  perpendicular  to  a  line  can  be  drawn  from  a  point  in 
the  line. 

Because  the  right  angles  would  not  be  equal  if  there 
were  two  perpendiculars  ;  and  all  right  angles  are  equal. 
(See  42.) 

18 


BOOK  i  19 

44.  If  two  adjacent  angles  have  their  exterior  sides  in  a  straight 
line,  they  are  supplementary. 

Because  they  together  =  two 
rt.  A.     (See  20.) 

45.  If  two  adjacent  angles  are 

supplementary,  their  exterior  sides 

Are  in  the  same  straight  line. 

Because  their  sum  is  two  rt.  A  (20);  or  a  straight  Z 
(37).  Hence  the  exterior  sides  are  in  the  same  straight 
line  (18). 

46.  The  sum  of  all  the  angles  on  one  side  of  a  straight  line  at  a 
point  equals  two  right  angles. 

(See  Ax.  4  and  37.) 

47.  The  sum  of  all  the  angles  about 
a  point  in  a  plane  is  equal  to  four 
right  angles.     (See  46.) 

48.  Angles  that  have  the  same  complement  are  equal.    Or,  comple- 
ments of  the  same  angle,  or  of  equal  angles,  are  equal. 

Because  equal  angles  subtracted  from  equal  right  angles 
leave  equals.  (See  Ax.  2.) 

49.  Angles  that  have  the  same  supplement  are  equal.    Or,  supple- 
ments of  the  same  angle,  or  of  equal  angles,  are  equal.     (  See  Ax.  2.) 

50.  If  two  angles  are  equal  and  supplementary,  they  are  right 
angles. 

Because  each  is  half  a  straight  Z ;  hence  each  is  a  rt.  Z. 
(See  36.) 

NOTE.  A  single  number,  given  as  a  reference,  always  signifies  the 
truth  stated  in  that  paragraph  and  is  usually  the  statement  in  full  face 
type  only.  In  reciting  or  writing  the  demonstrations  the  pupil  should 
quote  the  correct  reason  for  each  statement,  and  not  give  the  number 
of  its  paragraph.  [Consult  model  demonstrations  on  page  24.] 


20  PLANE  GEOMETRY 

THEOREMS  AND  DEMONSTRATIONS 

51.   THEOREM.     Vertical  angles  are  equal. 

Given:    A AOM    and   BOL,   a 
pair  of  vertical  angles. 

To  Prove:  . 


Proof:  Z.AOM  is  the  supple- 
ment of  Z.  MOB.  (Why?) 
(See  44.) 

Z.BOL  is  the  supplement  of  /.MOB.     (Why?)   (See  44.) 

/.  ^AOM=/:  BOL.     (Why  ?)  (See  49.) 

A  A  OL  and  BOM  are  a  pair  of  vertical  angles.  These  may  be  proved 
equal  in  precisely  the  same  manner.  If  Z  AOL  =  80°,  how  many  degrees 
are  there  in  the  other  A  ? 

52.  THEOREM.  Two  triangles  are  equal  if  two  sides  and  the  in- 
cluded angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

c 


B  R 


Given:     &  ABC  and RST;  AB  =  ES\  AC  =  RT\  Z.  A  =  Z  R. 
To  Prove  :     A  ABC  =  A  RST. 

Proof:     Place  the  A  ABC  upon  the  A  RST  so  that  Z  A  co- 
incides with  its  equal,  Z  R ;    then  AB  will  fall  upon  RS  and 
point  B  upon  s.     (It  is  given  that  AB  =  RS.)     AC  will  fall 
upon  RT  and  point  C  upon  T.     (It  is  given  that  AC  =  RT.) 
/.  BC  will  coincide  with  ST.     (Why?)  (See  39.) 
Hence,  the  triangles  coincide  in  every  respect   and   are 
equal  (28). 


BOOK  I 


21 


53.  THEOREM.    Two  right  triangles  are  equal  if  the  two  legs  of 
one  are  equal  respectively  to  the  two  legs  of  the  other. 

Given:  Rt.  &ABC&ndDEF; 
AC  =  DF\  CB  =  FE. 

•   To  Prove :   A  ABC  =  A  DEF. 

Proof :  In  the  A  ABC  and 
DEF,  AC  =  DF  (Given)  ;  CB 
=  FE  (Given)  ;  Z  C  =  Z  F. 
(Why?)  (See  42.)' 

/.  the  A  are  equal.  (Why?) 
(Theorem  of  52.) 

54.  THEOREM.    Two  triangles  are  equal  if  a  side  and  the  two 
angles  adjoining  it  in  the  one  are  equal  respectively  to  a  side  and 
the  two  angles  adjoining  it  in  the  other. 


Given:    A  BCD  and  JKL  ;     BC  =  JK\    Z  £  =  Z  J ;     Z  C  = 
ZJT. 

To  Prove  :    A  BCD  =  A  JKL. 

Proof:    Place  A  BCD  upon  A  JKL  so  that  Z  B  coincides 

with  its  equal,  Z  J,  BC  falling  on  JK. 

Point  C  will  fall  on  K.     (It  is  given  that  BC=JK.~) 
BD  will  fall  on  JL.     (Because  Z  B  is  given  =  Z  ,7.) 
CD  will  fall  on  KL.     (Because  Z  C  is  given  =  Z  K.) 
Then  point  D  which  falls  on  both  the  lines  JL  and  KL  will 

fall  at  their  intersection,  L.    (Why  ?)     (See  38.) 
/.  the  A  are  =.     (Why?)     (See  28.) 


22 


PLANE   GEOMETRY 


55.    THEOREM.    The  angles  opposite  the  equal  sides  of  an  isosceles 
triangle  are  equal. 

Given  :      A  ABC,  AB  —  AC. 
To  Prove  :     Z  B  =  Z  c. 

Proof:  Suppose  AX  is 
drawn  dividing  Z  BAG  into 
two  equal  angles,  and  meeting 
BC  at  X.  In  A  BAX  and 
CAX,  AX—  AX  (Identical); 
AB=AC  (Given);  Z.  BAX 


(Because  AX  made  them  =  .)    .-.  A  ABX=A  ACX. 
(Why?)  (52.)     .-.ZJ8  =  ZC.   (Why?)   (See  27.) 

56.  THEOREM.    An  equilateral  triangle  is  equiangular.   (See  55.) 

57.  THEOREM.    The  line  bisecting  the  vertex-angle  of  an  isosceles 
triangle  is  perpendicular  to  the  base,  and  bisects  the  base. 

Prove  A  ABX  and  ACX  equal  as  in  55.  Then,  Z  AXB 
=  Z  AXC.  (Why?)  (27.)  .'.  A  AXB  and  AXC&YQ  rt.  A  (16). 

.\AX  is  J_  to  BC.  (Why?)  (17.)  And,  also,  BX=  CX. 
(Why?)  (27.) 

58.  THEOREM.    Two  triangles  are  equal,  if^  the  three  sides  of  one 
are  equal  respectively  to  the  three  sides  of  the  other. 


Given:   A  ABC  and  RST\  AB  — 

C=ET;   BC=ST. 

To  Prove :  A  EST=A  ABC. 


15 


BOOK  I  23 

Proof :     Place  A  ABC  in  the  position  of  A  AST  so  that  the 
longest  equal  sides  (BC  and  8T)  coincide  and  A  is  opposite 
ST  from  R.     Draw  RA.      RS  =  AS  (Given).     /.  ASR  is  an 
isosceles  A.     (Def.  24.)' 
.  Z  8RA  ==  Z  SAR.  (Why?)  (55.)  Likewise  TR  =  AT  (?)  and 

Z  TEA  =  Z  TAR.  (Why?)    Adding  these  equals  we  obtain 

Z  SRT  =  /.  SAT   (Ax.  2).      /.  ARST=A  AST  (52). 

That  is,  A  RST  =  A  ABC.     (Substitution,  Ax.  6.) 

59.  Elements  of  a  theorem.      Every  theorem  contains  two 
parts,  the  one  is  assumed  to  be  true  and  the  other  results 
from  this  assumption.     The  one  part  contains  the  given  con- 
ditions, the  other  part  states  the  resulting  truth. 

The  assumed  part  of  a  theorem  is  called  the  hypothesis. 

The  part  whose  truth  is  to  be  proved  is  the  conclusion. 

Usually  the  hypothesis  is  a  clause  introduced  by  the  word 
"if."  When  this  conjunction  is  omitted,  the  subject  of  the 
sentence  is  known  and  its  qualities,  described  in  the  quali- 
fying words,  constitute  the  "given  conditions."  Thus,  in 
the  theorem  of  58,  the  assumed  part  follows  the  word  "  if," 
and  the  truth  to  be  proved  is:  "  Two  triangles  are  equal." 

60.  Elements  of  a  demonstration.     All  correct  demonstra- 
tions should  consist  of  certain  distinct  parts,  namely : 

1.  Full  statement  of  the  given  conditions  as  applied  to  a 
particular  figure. 

2.  Full  statement  of  the  truth  which  it  is  required  to  prove. 

3.  The  Proof.    This  consists  in  a  series  of  successive  state- 
ments, for  each  of  which  a  valid  reason  should  be  quoted. 
(The  drawing  of  auxiliary  lines  is  sometimes  essential,  but 
this    part  is    accomplished    by    imperatives    for    which    no 
reasons  are  necessary.) 

4.  The  conclusion  declared  to  be  true. 

The  letters  "Q.E.D."  are  often  annexed  at  the  end  of  a 
demonstration  and  stand  for  "  quod  erat  demonstrandum," 
which  means,  "  which  was  to  be  proved." 


PLANE   GEOMETRY 


MODEL    DEMONSTRATIONS 


The  angles  opposite  the  equal  sides  of  an  isosceles  triangle  are  equal. 

Given:   &ABC;  AB  =  AC. 

To  Prove :  Z  B  =  Z  C. 

Proof  :     Suppose    A  X    is    drawn    bisecting 
Z.BAC  and  meeting  BC  at  X. 

In  the  A  BAX  and  CAX 
AX  =  AX  (Identical). 
AB  =  AC  (Hypothesis). 

ZBAX  =  Z  CAX  (Construction). 

.\&ABX  -  &ACX.  (Two  A  are  =  if  two  sides  and  the  included  Z  of 
one  are  —  respectively  to  two  sides  and  the  included  Z  of  the  other.) 

Hence,  Z  B  =  Z  C.  (Homologous  parts  of  equal  figures  are  equal.)  Q.E.D. 


B 


Two  triangles  are  equal  if  the  three  sides  of  one  are  equal  respectively  to 
the  three  sides  of  the  other. 

Given  :  &  ABC  and 
RST;  AB  =  RS;  AC= 
RT;  BC  =  ST. 

To  Prove  :  A  RST  = 
&ABC. 

Proof:  Place  A  ABC  in 
the  position  of  A  AST  so 
that    the    longest    equal 
sides  {BC  and  ST)  coincide,  and  A  is  opposite  ST from  R.    Draw  RA. 
RS  =  AS  (Hypothesis). 

A  ASR  is  isosceles.  (An  isosceles  A  is  a  A  two  sides  of  which  are  equal.) 
.*.  Z  SRA  =  Z  SAR  . . .  (1)  . .  (The  A  opp.  the  =  sides  of  an  isos.  A  are  = .) 

Again,  TR  =  A  T  (Hypothesis). 
A  TRA  is  isosceles.  (Same  reason  as  before.) 

Z  TRA  =  Z  TAR . . .  (2)  .  .  (Same  reason  as  for  (1).)  Adding  equations 

(1)  and  (2). 

Z  SRT  =  Z  SA  T.  (If  ='s  are  added  to  ='s  the  results  are  =.) 

Consequently,  the  A  RST  =  A  AST.  (Two  A  are  =  if  two  sides  and 
the  included  Z  of  one  are  =  respectively  to  two  sides  and  the  included 
Z  of  the  other.) 

That  is,  &RST  =  &ABC.     (Substitution;   A  ABC  is  the  same  as 

&AST.)  Q.E.D. 


BOOK  I  25 

The  preceding  form  of  demonstration  will  serve  to  illustrate  an  excel- 
lent scheme  of  writing  the  proofs.  It  will  be  observed  that  the  statements 
are  at  the  left  of  the  page  and  their  reasons  at  the  right.  This  arrange- 
ment will  be  found  of  great  value  in  the  saving  of  time,  both  for  the 
pupil  who  writes  the  proofs  and  for  the  teacher  who  reads  them. 

61.  The  converse  of  a  theorem  is  the  theorem  obtained  by 
interchanging  the  hypothesis  and  conclusion  of  the  original 
theorem.     Consult  44  and  45;  79,  80,  and  others. 

Every  theorem  which  has  a  simple  hypothesis  and  a  simple 
conclusion  has  a  converse,  but  only  a  few  of  these  converses 
are  actually  true  theorems. 

For  example  :  Direct  theorem  :  "  Vertical  angles  are 
equal." 

Converse  theorem  :  "If  angles  are  equal,  they  are  verti- 
cal." This  statement  cannot  be  universally  true. 

The  theorem  of  120  is  the  converse  of  that  of  55. 

62.  Auxiliary  lines.    Often  it  is  impossible  to  give  a  simple 
demonstration  without  drawing  a  line  (or  lines)  not  described 
in  the  hypothesis.    Such  lines  are  usually  dotted  for  no  other 
reason  than  to  aid  the  learner  in  distinguishing   the  lines 
mentioned  in  the  hypothesis  and  conclusion  from  lines  whose 
use  is  confined  to  the  proof.     Hence,  lines  mentioned  in  the 
hypothesis  and  conclusion  should   never  be  dotted.    (The 
figure  used  in  57  should  contain  no  dotted  line.) 

63.  Converse  of  definitions.     The  converse  of  a  definition 
is  true.     It  is  often  advantageous  to  quote  the  converse  of 
a  definition,  as  a  reason,  instead  of  the  definition  itself. 

64.  Homologous  parts.   Triangles  are  proved  equal  in  order 
that  their  homologous  sides,  or  homologous  angles,  may  be 
proved  equal.     This  is  a  very  common  method  of  proving 
lines  equal  and  angles  equal. 

65.  The  distance  from  one  point  to  another  is  the  length 
of  the  straight  line  joining  the  two  points. 


26 


PLANE   GEOMETRY 


66.  THEOREM.    If  lines  be  drawn  from  any  point  in  a  perpendicu- 
lar erected  at  the  midpoint  of  a  straight  line  to  the  ends  of  the  line, 

I.    They  will  be  equal. 

II.    They  will  make  equal  angles  with  the  perpendicular. 
III.    They  will  make  equal  angles  with  the  line. 

Given  :  AB  _L  to  CD  at  its 
midpoint,  £;  P  any  point  in 
AB  ;  PC  and  PD. 

To  Prove:  I.  PC=PD; 

II.  Z  CPB  =  Z  DPS  ;   and 
III.  Z  C  =  Z  D. 

Proof:    Iii   rt.  A   P-BC  and    c< 
PJ3D,  BC  =  BD  (Hyp.)  ;  BP  =  BP  (Iden.). 
.•.AP£C  =  APJ3D.    (Why?)    (53.) 

.-.   I.   PC  =  PD(Why?)(27;)  II.  Z  CPJ3  =  Z  DPS  ( Why  ?) ; 

III.  Zc  =  ZD  (Why?).  Q.E.D. 

67.  THEOREM.  Any  point  in  the  perpendicular  bisector  of  a  line  is 
equally  distant  from  the  extremities  of  the  line.     (See  66,  I.) 

68.  THEOREM.     Any  point  not  in  the  perpendicular  bisector  of  a 
line  is  not  equally  distant  from  the  extremities  of  the  line. 

Given :  AB  _L  bisector  of  CD ;  A 

P  any  point   not   in   AB ;    PC 
and  PD.  o 

To  Prove :  PC  not  =  PD. 

Proof  :    Either  PC  or  PD  will 
cut  AB. 

Suppose    PC  cuts   AB   at  O. 
Draw  OD. 

DO  +  OP  >  PD.    (Why  ?)  (Ax.  12.)    But  CO  =  OD  (67). 

.'.  CO+  OP  >  PD.  (Substitution;  Ax.  6.) 

That  is,  PC  >  PD,  or  PC  is  not  =  PD.  Q.E.D. 


BOOK  I  27 

69.  THEOREM.   If  a  point  is  equally  distant  from  the  extremities  of 
a  line,  it  is  in  the  perpendicular  bisector  of  the  line.  (See  67  and  68.) 

70.  THEOREM.   Two  points  each  equally  distant  from  the  extrem- 
ities of  a  line  determine  the  perpendicular  bisector  of  the  line. 

Each  point  is  in  the  _L  bisector  (69) ;   two  points  deter- 
mine a  line  (5). 

71.  THEOREM.   Only  one  perpendicular  can  be  drawn  to  a  line  from 
an  external  point. 

Given:  PR  J_  to  AB  from  P; 
PD  any  other  line  from  P  to  AB. 

To  Prove :  PD  cannot  be  J_  to 
AB ;  that  is,  PR  is  the  only  _L  to 
AB  from  P. 

Proof:  Extend  PR  to  8,  mak- 
ing RS  =  PR  ;  draw  DS. 

In  rt.  A  PDR  and  SDR,  PR  =  R8 
(Const.). 

DR  =  DR  (Iden. ).  . ' .  A  PDR  = 
A  SDR.  (Why?)  (53.) 

.'.  Z  PDR  =  Z  SDR  (27).    That  is,  Z  PDR  =  half  of  Z  PDS. 

Now  PRS  is  a  straight  line  (Const.). 

.'.PDS  is  not  a  straight  line  (39). 

.*.  Z  PDS  is  not  a  straight  angle  (18). 

/.  /.PDR,  the  half  of  Z  PDS,  is  not  a  right  angle  (36). 

.*.  PD  is  not  _L  (17).     /.  PR  is  the  only  _L.  Q.E.D. 


Ex.  1.  Through  how  many  degrees  does  the  minute  hand  of  a  clock 
move  in  15  min.  ?  in  20  min.  ?  Through  how  many  degrees  does  the  hour 
hand  move  in  one  hour?  in  45  minutes?  in  10  minutes? 

Ex.  2.  How  many  degrees  are  there  in  the  angle  between  the  hands 
of  a  clock  at  9  o'clock  ?  at  10  o'clock  ?  at  12  : 30  ?  at  2  : 15  ?  at  3  : 45  ? 

Ex.  3.  THEOREM.  If  two  lines  be  drawn  bisecting  each  other,  and 
their  ends  be  joined  in  order,  the  opposite  pairs  of  triangles  will  be 
equal.  [Use  51  and  52.] 


28  PLANE   GEOMETRY 

72.  THEOREM.  Two  right  triangles  are  equal  if  the  hypotenuse 
and  an  adjoining  angle  of  one  are  equal  respectively  to  the  hypote- 
nuse and  an  adjoining  angle  of  the  other. 

N  T 


L  MR  S 

Given:     Rt.  A  LMN  and  EST  ;  LN  =  BT  ;  and  Z  L  =  /.E. 
To  Prove:    A  LMN  =  A  EST. 

Proof  :  Superpose  A  LMN  upon  A  EST  so  that  Z  L  coincides 
with  its  equal,  Z  E,  LM  falling  along  ES.  Then  LN  will  fall 
on  ET  and  point  N  will  fall  exactly  on  r  (LN—  ET  by  Hyp.). 

Now  jyjf  and  TS  will  both  be  J_  to  ES  from  r  (Rt.  A 
by  Hyp.).  .*.  N  M  will  coincide  with  TS  (71). 

/.  A  LMN  =  A  KSr  (28).  Q.E.D. 


73.  THEOREM.  Two  right  triangles  are  equal  if  the  hypotenuse 
and  a  leg  of  one  are  equal  respectively  to  the  hypotenuse  and  a  leg 
of  the  other. 

K  R 


I*"  —  J  X  L 


Given  :    Rt.  A  UK  and  iJf£  ;  KI  =  EM  ;  KJ  =  EL. 
To  Prove  :    A  UK  =  A  LME. 

Proof  :  Place  A  UK  in  the  position  of  A  XLE  so  that 
the  equal  sides,  KJ  and  EL,  coincide  and  I  is  at  X,  oppo- 
site Ei  from  M. 


BOOK  I 


29 


Now,  A  ELM  and  ELX  are  supplementary.    (Why  ?)  (20.) 

.'.  XLM  is  a  str.  line  (45). 

Also,  A  EMX  is  isosceles.  (Why  ?)  (EX  =  EM  by  Hyp.) 
,\ZX=ZJf.  (Why?)  (55.) 

/.  A  XLE  =  A  JfLtf.  (Why?)  (72.)  That  is,  A  UK  = 
ALMK.  (Ax.  6.)  Q.E.D. 

COR.  The  perpendicular  from  the  vertex  of  an  isosceles  triangle  to 
the  base  bisects  the  base. 

Proof  :   A  XLR  =  A  MLR.     (Why  ?)      /.  XL  =  ML.     (Why  ?) 

74.  THEOREM.  Two  right  triangles  are  equal  if  a  leg  and  the  ad- 
joining acute  angle  of  one  are  equal  respectively  to  a  leg  and  the 
adjoining  acute  angle  of  the  other. 


Given  :    Rt.  A  ABC  and  DEF  ;  AC  =  DF  ; 
To  Prove  :    A  ABC  =  A 


=  Z  D. 


Proof  :  In  the  A  ABC  and  DEF,  AC  =  DF.  (Why?)  (Hyp.) 
Also  Z.A  =  Z.D  (Why?)  and  ZC  =  Z.F.  (Why?)  (42.) 
.'.A  ABC  =  A  DEF.  (Why?)  (54.)  Q.E.D. 


Ex.  1.  How  many  pairs  of  equal  parts  must  two  triangles  have,  in 
order  that  they  may  be  proved  equal?  How  many  pairs  is  it  necessary 
to  mention  in  the  case  of  two  right  triangles? 

Ex.  2.  THEOREM.  If  a  perpendicular  be  erected  at  any  point  in  the 
bisector  of  an  angle,  two  equal  right  triangles  will  be  formed.  [Use  74.] 

Ex.   3.    Through  the  midpoint  of  a  line  AB  any  oblique  line  is  drawn  : 

I.   The  lines  JL  to  it  from  A  and  B  are  equal.     [Use  72.] 
II.   The  lines  _L  to  AB  at  A  and  B,  terminated  by  the  oblique  line, 
are  equal.     [Use  74.] 


so 


PLANE  GEOMETRY 


75.  THEOREM.  The  sum  of  two  sides  of  a  triangle  is  greater  thaii 
the  sum  of  two  lines  drawn  to  the  extremities  of  the  third  side,  from 
any  point  within  the  triangle. 

Given:    P,  any  point  in 
A  ABC ;  lines  PA    and  PC. 
To  Prove :    AE    +  BC  > 
AP  +  PC. 

Proof :    Extend     AP    to 
meet  BC  at   x.  A 

AB  +  BX  >  AP  +  PX.  (Why  V)  (Ax.  12.) 

CX  +  PX  >  PC. (Why?)  (Ax.  12.)  Add  : 


AB 


BX  +  CX  +  PX  >  AP  +  PC  +  Px(Ax.  8). 
Subtract  PJT  =  PX. 

.^LB  +  SC  >  AP  +  PC  (Ax.  7).  Q.E.D. 


76.     THEOREM.     If  from  any  point  in  a  perpendicular  to  a  line 
two  oblique  lines  be  drawn, 

I.   Oblique  lines  cutting  off  equal  distances  from  the  foot  of  the 
perpendicular  will  be  equal. 

II.    Equal  oblique  lines  will  cut  off  equal  distances  (converse). 
III.    Oblique  lines  cutting  off  unequal  distances  will  be  unequal,  and 
that  one  which  cuts  off  the  greater  distance  will  be  the  greater. 


I.    Given  :    CD  _L  to  AB  ;    ND  =  MD  ; 
oblique  lines  PN  and  PM.   [First  figure.] 

To  Prove  :   PN  =  PM. 


BOOK  I 


31 


Proof  :    PD  is  JL bisector  of  NM  (Hyp.).    .-.  PN  =  PM  (67). 

II.  Given:    CD  _L  to  AB  ;  PN  =  PM.     [First  figure.] 
K    To  Prove  :   ND  =  MD. 

Proof:    In   the  rt.  A    PND    and  PM D,   PD  =  PD  (Iden.); 

and  PN  =  PM  (Hyp.).    .-.  APND  =  APMD.   (Why?)  (73.) 

.\ND=MD.      (Why?)      (27.)  Q.B.D. 

III.  Given :    CD  _L  to  AB  ;    oblique   lines   PR,   PT ;  also 
RD  >  DT.     [Second  figure.] 

To  Prove:    PR  >  PT. 

Proof  :  Because  RD  is  >  DT,  we  may  mark  z>s  (on  RD)  = 
Dr.  Draw  PS.  Extend  PD  to  JT,  making  DX  =  PD ; 
draw  RX  and  &Z.  ^D  is  JL  to  PX  at  its  midpoint  (Const.). 
.'.PR=  RX  and  PS  =  SX  (66). 

Now  PR  +  RX  >  PS  +  sx  (75). 

Hence  P#  +  P#  >  PS  +  PS  (Ax.  6). 

That  is,  2  PE  >  2  PS.    /.  P#  >  PS  (Ax.  10). 

But  P5=  PT  (76,  I).     .*.  PU  >  PT  (Ax.  6).  Q.E.D. 

COR.  Therefore,  from  an  external  point  it  is  not  possible  to  draw 
three  equal  lines  to  a  given  straight  line. 

77.  THEOREM.  The  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  point  to  a  straight  line. 

Given :  PR  JL  to  AB ;  PC  not  _L. 
To  Prove :  PR  <  PC. 

Proof :  Extend  PE  to  X,  mak- 
ing RX  s=  PR.  Draw  CX.  PR  + 
RX  <  PC+  CX  (Ax.  12). 

But  AR  is  _L  to  PXat  its  mid-     A~~ 
point  (Const.). 

/.  PC=cx  (66). 

.'.  PR H-  PR  <  PC+ PC  (Ax.  6). 

That  is,  2 PR  <  2  PC. 

.'.  PR<  PC  (Ax.  10).     Q.E.D. 


N 


32  PLANE   GEOMETRY 

78.  The  distance  from  a  point  to  a  line  is  the  length  of  the 
perpendicular  from  the  point  to  the  line.     Thus  "  distance 
from  a  line  "  involves  the  perpendicular.     If  the  perpendicu- 
lars from  a  point  to  two  lines  are  equal,  the  point  is  said  to  be 
equally  distant  from  the  lines. 

79.  THEOREM.    Every  point  in  the  bisector  of  an  angle  is  equally 
distant  from  the  sides  of  the  angle. 

Given :    Z  ACE;    bisector    CQ; 
point  P  in  CQ;  distances  PB  and 


To  Prove :    PB  =  PD. 

Proof :    A  P.BC  and    PDC   are 

rt.  A  (78).  ~D~ 

In  rt.  A  PBC  and  PDC,  PC  =PC  (Iden.) ;    Z  PCB  =  Z  PCD 
(Hyp.).    /.APBC=APDC(?)(72).    .-.  PB  =  PD  (?).  Q.E.D. 

80.    THEOREM.    Every  point  equally  distant  from  the  sides  of  an 
angle  is  in  the  bisector  of  the  angle. 

Given :    Z  ACE;  P  a  point,  such  that  PB  =  PD  (distances); 
CQ  a  line  from  vertex  of  the  angle,  and  containing  P. 

To  Prove  :    Z  ^ICQ  =  Z  ECQ. 

Proof:    A  P.BC  and  PDC  are  right  A  (?).     In  rt.  A  PBC 
and  PDC,  PC  =  PC  (?)  ;  PB  =  PD  (?) . 

0(73).       .-.Z^4CQ  =  ZJ£CQ(?).    Q.E.D. 


81.  THEOREM.    Any  point  not  in  the  bisector  of  an  angle  is  not 
equally  distant  from  the  sides  of  the  angle.     [Because   if  it  were 
equally  distant,  it  would  be  in  the  bisector    (80).] 

82.  THEOREM.    The  vertex  of  an  angle  and  a  point  equally  distant 
from  its  sides  determine  the  bisector  of  the  angle.     (See  80  and  5.) 


Ex.   Describe  the  path  of  a  moving  point  which  shall  be  equally  dis- 
tant from  two  intersecting  lines. 


BOOK  I  33 

83.  The  altitude  of  a  triangle  is  the  perpendicular  from 
any  vertex  to  the    opposite  side  (prolonged  if  necessary). 
A  triangle  has  three  altitudes.     The  bisec- 
tor of  an  angle  of  a  triangle  is  the  line 

dividing   any   angle    into    equal    angles. 
A  triangle  has  three  bisectors  of  its  an- 

rrn  ,.  £         ,     •  T       .       ,1         T  THE    THREE    MEDIANS 

gles.     Ihe  median  01  a  triangle  is  the  line 

drawn  from  any  vertex  to  the  midpoint  of  the  opposite  side. 

A  triangle  has  three  medians. 

84.  THEOREM.   The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
point  which  is  equally  distant  from  the  sides. 

Given :  A  ABC,  AX  bisect- 
ing Z.  A,  B  T  and  CZ  the  other 
bisectors. 

To  Prove:   AX,     BY,     CZ 

meet  in  a  point  equally  dis- 
tant from  AB,  AC,  and  BC. 

Proof:  Suppose  that  AX 
and  BY  intersect  at  O. 

O  in  AX  is  equally  distant  from  AB  and  AC  (?)  (79). 

O  in  BY  is  equally  distant  from  AB  and  BC  (?). 

.•.  point  O  is  equally  distant  from  AC  and  BC   (Ax.  1).* 

.-.  O  is  in  bisector  CZ  (?)  (80). 

That  is,  all  three  bisectors  meet  at  O,  and  O  is  equally  dis- 
tant from  the  three  sides.  Q.E.D. 
*  The  JL  distances  from  0  to  the  three  sides  are  the  three  equals. 


Ex.  1.  Draw  the  three  altitudes  of  an  acute  triangle;  of  an  obtuse 
triangle. 

Ex.  2.   Prove  that  in  an  equilateral  triangle: 

(a)    An  altitude  is  also  a  median.     [Use  73.] 

(6)    A  median  is  also  an  altitude.     [Use  58,  27,  16,  17.] 

(c)  An  altitude  is  also  the  bisector  of  an  angle  of  the  triangle. 

(d)  'The  bisector  of  an  angle  is  also  an  altitude.     [Use  52.] 

(e)  The  bisector  of  an  angle  is  also  a  median. 


34 


PLANE   GEOMETRY 


M 


85.  THEOREM.   The  three  perpendicular  bisectors  of  the  sides  of  a 
triangle  meet  in  a  point  which  is  equally  distant  from  the  vertices. 

Given:  A  ABC;  LR,  MS,  NT, 
the  three  _L  bisectors. 

To  Prove :  LR,  MS,  NT  meet 
at  a  point  equally  distant  from 
A  and  B  and  C. 

Proof :  Suppose  that  LR  and 
MS  intersect  at  O. 

O  in  LR  is  equally  distant 
from^i  and  B.  (?)  (67.) 

O  in  MS  is  equally  distant 
from  A  and  C.  (?)  ~~N~~ 

.*.  point  O  is  equally  distant  from  B  and  C  (Ax.  1).* 

Hence  O  is  in  J.  bisector  NT  (?)  (69). 

That  is,  all  three  _L  bisectors  meet  at  O,  and  O  is  equally 
distant  from  A  and  B  and  C.  Q.E.D. 

*  The  three  lines  from  0  to  the  vertices  are  the  three  equals. 

86.  THEOREM.   If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  included  angle  in  the  first  greater  than  the 
included  angle  in  the  second,  the  third  side  of  the  first  is  greater  than 
the  third  side  of  the  second. 


Given:   A  ABC,  DEF;   AB  =  DE;  BC  =  EF;  Z  ABC>  Z  E. 


BOOK  I 


35 


To  Prove  :   AC  >  DF. 


Proof  :  Place  the  A  DEF  upon  A  ABC  so  that  line  DE 
coincides  with  its  equal  AB,  A  DEF  taking  the  position  of 
A  ABH.  There  will  remain  an  angle,  HBC.  (Z  ABC  is  >  Z  E.) 

Suppose  BX  to  be  the  bisector  of  Z  HBC,  meeting  AC  at  x. 
Draw  XH. 

In  AtfBXand  CBX,  BX=BX  (?);  Bff=  £(?(?); 
ZCBX(?)  (Const.).   .'.&HBX  =  ACBX  (?)(52). 
XC  (?). 

Now,  Ar+xfl-  >  ^ff(?).    .-.4-r  +  xc  >  ^ifl  (Ax.  6). 

That  is,  AC  >  DF.  Q.E.D. 

87.  THEOREM.  If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other  but  the  third  side  of  the  first  greater  than  the  third 
side  of  the  second,  the  included  angle  of  the  first  is  greater  than  the 
included  angle  of  the  second.  [Converse.] 


Given :   A  ABC  and  RST;  AB  =  US  ;  BC  =  sr ;  -AC  >  #T. 

To  Prove:    Z  B  >Z&'. 

Proof  :    It  is   evident   that   Z  B  <  Z  -8,  or  Z  B  =  Z  5,  or 


1.  It  Z.B  <  Z.S,  AC  <  BT  (86). 

But  ^4(7  >  RT  (Hyp.)-     .*.Z#  is  W0£  <  Z  S. 

2.  If  Z  £  =  Z  5,  the  A  are  =  (52),  and  AC  is  =  J2T  (27). 
But  AC  >  RT  (Hyp.).     /.Z  B  is  no£  =  Z  -8. 

3.  /.  the  only  possibility  is  that  Z  B  >  Z.S.  Q.E.D. 


36  PLANE   GEOMETRY 

88.  The  preceding  method  of  demonstration  is  termed  the 
method  of  exclusion.     It    consists    in   making    all    possible 
suppositions,  leaving  the  probable  one  last,  and  then  proving 
all  these  suppositions  impossible,  except  the  last,  which  must 
necessarily  be  true. 

89.  The  method  of  proving  the  individual  steps  is  called 
reductio    ad   absurdum    (reduction  to  an  absurd  or  impos- 
sible conclusion).     This   method   consists   in    assuming   as 
false  the  truth  to  be   proved  and  then   showing  that  this 
assumption    leads    to    a   conclusion  altogether   contrary  to 
known  truth  or  the  given  hypothesis.    (Examine  87.)    This 
is  sometimes  called  the  indirect  method.     The  theorems  of 
93  and  94  are  demonstrated  by  a  single  use  of  this  method. 

90.  THEOREM.    If  two  unequal  oblique  lines  be  drawn  from  any  point 
in  a  perpendicular  to  a  line,  they  will  cut  off  unequal  distances  from  the 
foot  of  the  perpendicular,  and  the  longer  oblique  line  will  cut  off  the  greatei 
distance.    [Converse  of  76,  III.]  C 

Given:  CDj_toAB;  PR  and. 
PS  oblique  lines  ;  PR  >  PS. 

To  Prove  :   DR  >  DS. 

Proof:  It  is  evident  that 
DR  <  DS,  or  DR  =  DS,  or 
DR  >  DS.  A~ 

If  DR  <  DS,  PR  <  PS  (76,  III).    But  PR  >  PS  (Hyp.). 

.*.  DR  is  not  <  DS. 

If  DR  =  DS,  PR  =  PS  (76,  I).     But  PR  >  PS  (Hyp.). 

/.  DR  is  not  =  DS. 

Therefore  the  only  possibility  is  that  DR  >  DS.      Q.E.D. 

91.  Parallel  lines  are  straight  lines  that  lie  in  the  same  plane 
and  that  never  meet,  however  far  extended  in  either  direction. 

92.  AXIOM.   Only  one  line  can  be  drawn  through  a  point  parallel  to 
a  given  line. 


BOOK  I  37 

93.  THEOREM.  Two  lines  in  the  same  plane  and  perpendicular  to 
the  same  line  are  parallel. 

Given  :"  CD  and  EF  in  same 
plane  and  both  _L  to  AB. 

To  Prove :   CD  and  EF  \\  .       E 

Proof :  If  CD  and  EF  were  not  II,  they  would  meet  if  suffi- 
ciently prolonged.  Then  there  would  be  two  lines  from  the 
point  of  meeting  _L  to  AB.  (By  Hyp.  they  are  ±  to  AB.) 

But  this  is  impossible  (?)   (71). 

.-.CD  and  EF  do  not  meet,  and  are  parallel  (91).    Q.E.D. 

94.  THEOREM.  Two  lines  in  the  same  plane  and  parallel  to  the 
same  line  are  parallel. 

Given :  AB  \\  to  RS  and  CD  \\  to  RS  and  in  the  same  plane. 
To  Prove :  AB  \\  to  CD.  A B 

Proof :  If  AB  and  CD  were 
not  ||,  they  would  meet  if 
sufficiently  prolonged. 

Then  there  would  be  two  lines  through  the  point  of  meet- 
ing ||  to  the  line  RS.  (By  Hyp.  they  are  ||  to  RS). 

But  this  is  impossible  (92). 

/.  AB  and  CD  do  not  meet,  and  are  ||  (?)  (91).          Q.E.D. 

95.  THEOREM.  If  a  line  is  perpendicular  to  one  of  two  parallels,  it  is 
perpendicular  to  the  other  also. 

Given :  LM  _L  to  AB  and 
AB  ||  to  CD. 


To  Prove :  LM  _L  to  CD.       X -g — • 

Proof:   Suppose    XY   is 
drawn  through  M  _L  to  LM.     AB  is  ||  to  XY  (?)  (93). 
But  AB  is  ||  to   CD  (Hyp.). 
Now  CD  and  XY  both  contain  M  (Const.). 
.'.CD  and  XY  coincide   (92). 
But  LM  is  _L  to  XY  (?).   That  is,  LM  is  J_  to  CD.          Q.E.D. 


38 


PLANE   GEOMETRY 


96.  If  one  line  cuts  other  lines,  it  is  called  a  transversal. 
Angles  are  formed  at  the  several  intersections,  and   these 
receive  the  following  names  : 

Interior  A  are  between  the  lines  [6,  c,  e,  K], 
Exterior  A  are  without  the  lines  [a,  d,f,  g~\. 
Alternate  A  are  on   opposite  sides  of  the 
transversal  [b  and  h  ;  c  and  e  ;  a  and  g ;  etc.]. 

Alternate-Interior  A   are  b  and  h;   and  c 
and  e. 

Alternate-Exterior  A  are  a  and  g ;  d  and/. 

Corresponding  A  are  a  and  e ;  d  and  A  ;  b  and/;  c  and  #. 

Adjoining-Interior  A  are  c  and  h ;  6  and  e. 

Adjacent-Interior  A  are  &  and  c ;  e  and  ft. 

The  left-hand  transversal  makes  eight  angles  similarly  related.  The 
"primes"  are  used  only  to  designate  angles  which  are  different  from 
those  in  the  right-hand  part  of  the  figure. 

97.  THEOREM.   If  a  transversal  intersects  two  parallels,  the  alter- 
nate interior  angles  are  equal. 

Given :    AB 
and  K. 


to  CD\    transversal  EF  cutting  the  Us  at  H 


To  Prove :     Z  a  =  Z  i  and 


Proof :  Suppose  through 
Jf,  the  midpoint  of  HE",  ES  is 
drawn  J_  to  AB.  Then  ES  is 
_L  to  CD  (95).  In  rt.  A  EMH 
and  JOfS,  HJf  =  .Of  (Const.); 
Z  EMH=Z.  KMS  (?)  (51).  .'.A  EMH  =  A  JTAfS  (?)  (72). 


Z  re  is  the  supp.  of  Z  a  (?)  (44)  ;  j  < 
Z  v  is  the  supp.  of  Z  i  (?). 


(49). 

Q.E.D. 


EXERCISE.   If  Z.  a  =  70°,  in  the  figure  of  97,  how  many  degrees  are 
there  in  Zx?  inZi?  Zv?  Z.AHE1  Z.EHB1  +CKF'!   ZDKF1 


BOOK  I  39 

98.   THEOREM.   If  a  transversal  intersects  two  parallels,  the  corre- 
sponding angles  are  equal. 

Given  :  AB  II  to  CD ;  trans- 
versal EF  cutting  the  Us  and 
forming  the  8  A.  A~~" 


z 


To  Prove :    Z  s  =  Z  i ;    Z  c 
Proof:      Zs  =  Za      (51); 


Z  <?=Z  m  (?)  ;  Z  w  =  Z  r  (?).  /.Z  c=Z  r  (?).    Etc.  Q.E.D. 

99.  THEOREM.   If  a  transversal  intersects  two  parallels,  the  alter- 
nate-exterior angles  are  equal. 

Given:    (?).     To  Prove:    (?).     Proof:   Zc=Zr  (?);  and 
Zr  =  Zrc(?).     •  '-/.  c=  /Ln  (?).    Etc.  Q.E.D. 

100.  THEOREM.    If  a  transversal  intersects  two  parallels,  the  sum  of 
the  adjoining  interior  angles  equals  two  right  angles. 

Given  :  (?).    To  Prove  :    Z  a  +  ^  r  =  2  rt.  A.     Etc. 

Proof:    Z  a  +  Zm  =  2rt.^(?)  (46).    ButZm  =  Zr(?). 
/.Z  a  +  Zr=  2  rt.  A  (Ax.  6).    Etc.  Q.E.D. 

101.  THEOREM.   If  a  transversal  intersects  two  lines  and  the  alter- 
nate-interior angles  are  equal,  the  lines  are  parallel.  [Converse  of  97.] 

Given:  AB  and  CD  two 
lines  ;  transversal  EF  cutting 
them  at  H  and  irrespectively; 
Z  a  =  Z  HKD. 

To  Prove  :   CD  \\  to  AB. 

'R* 

Proof:  Through  K,  suppose 
ES  is  drawn  ||  to  AB.     Then 
(97).     But 


(Hyp.).  .'.Z.HKS  =  Z.HKD  (?).  /.  JTD  and  M 
coincide:  that  is,  CD  and  BS  are  the  same  line.  .'.  CD  is  || 
to  AB.  (Because  it  coincides  with  $S  which  is  II  to  41?)  .  Q.E.D, 


40  PLANE   GEOMETRY 

102.   THEOREM.  If  a  transversal  intersects  two  lines  and  the  corre- 
sponding angles  are  equal,  the  lines  are  parallel. 

Given :  AB  and  CD  cut  by 
transversal  EF\  Z  c  =  Z  r. 


To  Prove  :  AB  II  to  CD. 
Proof:    Ztf  =  Zm  (?);    Z<? 


Zr  (.    c- 
.'.AB  is  II  to  CD  (101).    Q.E.D. 

If  Z  a  were  given  =  Z  o  or         F 
Z  «  =  Z  i,  etc.,  the  proof  that  ^.B  is  II  to  CD  would  be  the  same. 

103.  THEOREM.   If  a  transversal  intersects  two  lines  and  the  alter- 
nate-exterior angles  are  equal,  the  lines  are  parallel. 

Given  :  AB  and  CD  cut  by  EF,  and  Z  c  =  Z  n. 
To  Prove  :  A  B  II  to  CD. 

Proof:  Zc  =  Zw  (?);  Zc  =  Zn(?).  .'.Zw  =  Zrc  (?). 
.-.  ABis  II  to  OD(?).  Etc.  Q.E.D. 

104.  THEOREM.    If  a  transversal  intersects  two  lines  and  the  sum 
of  the  adjoining-interior  angles  equals  two  right  angles,  the  lines  are 
parallel. 

Given  :  AB  and  CD  cut  by  EF  and  Z  a  -f-  Z  r  =  2  rt.  A. 

To  Prove  :  AB  II  to  CD. 

Proof:  Z  tf  +  Z<?=2  rt.^  (46);  Z  «  +  Zr  =  2  rt.  ^  (?). 
.-.Za  +  Z  <?=Za  +  Zr  (?).  HenceZ<?  =  Zr  (Ax.  2). 
.-.  AB  is  H  to  CD  (?)  (102).  Q.E.D. 

,,     r  Zaissupp.  of  Z  <?(?);]  x     ,QN    T^, 

Another  proof:  ,    .     )o;      .*.Zc?=  Zr  (?).  Etc. 

[  Zaissupp.of  Zr  (/).  j 

105.  THEOREM.    If  two  angles  have  their  sides  parallel  each  to  each, 
the  angles  are  equal  or  supplementary. 

I.    Given:  Za  and  Z  b  with  their  sides  II  each  to  each  and 
extending  in  the  same  directions  from  their  vertices. 
To  Prove  :  Z  a  =  Z  b. 
Proof  :   If  the  non-parallel  lines  do  not  intersect,  produce 


BOOK  I  41 

them  till  they  meet,  forming  Z  o.    Now, 
Z  a  =  Z  o.  (?)  (98).  Z  o  =  Z.  b  (?). 
.-.Za  =  Z6  (?).  Q.E.D. 

II.  Given :  Z  a  and  Z  0  with  their 
sides  II  each  to  each  and  extending  in 
opposite  directions  from  their  vertices. 

To  Prove:  Z  a  =  Z  c. 
Proof:    Za  =  Z£  (Proved  in  I); 
Z  6  =  Z  <?  (?). 

Hence  Z  a  =  Z  <?  (?).  Q.E.D, 

III.  Given :  Z  a  and  Z  <i  with  their  sides  II  each  to  each, 
but  one  pair  extending  in  the  same  direction,  the  other  pai* 
extending  in  opposite  directions  from  their  vertices. 

To  Prove :  Z  a  and  Z  <#  supplementary. 
Proof :   Z  5  and  Z  c?  are  supplementary  (44). 
Za  =  Z6(I).  .-.Z  aandZ  dare  supp.  (Ax.  6.).  Etc.  Q  B.D 

106.   THEOREM.   If  two  angles  have  their  sides  perpendicular  each  to 
each,  the  angles  are  equal  or  supplementary. 

I.  Given:   A  a  and  b  with 
sides  _L  each  to  each. 

To  Prove :    Z  a  =  Z  b. 

Proof:  At  B  suppose  BE  is 
drawn  JL  to  BC  and  BS  _L  to  AB. 
BE  is  II  to  FE  and  BS  is  II  to 
DE  (?)  (93). 

.-.Z  RBS  =  Z.b  (?)  (105). 

Now,  Z  a  is  the  comp.  of  Z  ABfi  (20),) /.Z  a=  Z  #J5S  (?). 
and  Z  E£S  is  the  comp.  of  Z  ^tB£  (?).    J  (48.) 

/.Za  =  Z6.     (Ax.  1.)  Q.E.D. 

II.  Given :    A  a  and  c  with  sides  _L  each  to  each. 
To  Prove  :    Z  a  and  Z  c  supplementary. 

Proof:    Z  5  and  Z  c?  are  supp.   (?);    Z  a  =  Z  5  (I). 

.-.  Z  a  and  Z  c  are  supp.  (Ax.  6).  Q.B.D. 


42 


PLANE  GEOMETRY 


107.  An  exterior  angle  of  a  triangle 
is  an  angle  formed  outside  the    tri- 
angle, between  one  side  of  the  tri- 
angle and  another    side    prolonged. 
[Z  ABX.~\ 

The    angles   within   the    triangle, 
at  the  other  vertices  are  the  opposite  c 
interior  angles.     [Z  A  and  Z  c.] 

108.  THEOREM.    An  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  opposite  interior  angles. 

Given:    A  ABC;    exterior   Z 
ABD. 

To  Prove :    Z  ABD  =  Z  A  4-  p 

Z  c. 

Proof:     Suppose    BP    to    be 
drawn  through  B  II  to  AC. 

Z  ABD  =  Z  ABP  +  Z  PBD  (Ax.  4). 


Q.E.D. 


Z  C  (?)  (98). 

/.  Z  ^LBD  =  Z  ^  +  Z  C   (Ax.  6). 


109.  THEOREM.   An  exterior  angle  of  a  triangle  is  greater  than 
either  of  "the  opposite  interior  angles.       (See  Ax.  5.) 

110.  THEOREM.   The  sum  of  the  angles  of  any  triangle  is  two  right 
angles ;  that  is,  180°. 

Given :   A  ABC. 

To  Prove :    Z  A  +  Z  B  + 

=  2  rt.  ^  =  180°. 


Proof  :    Prolong  AC  to  JT, 

making  the  ext.  Z  BCX.          A~~  "~  c""" 

Z  BCX-}-  Z  ACB  ==  2  rt.  A  (?)  (46). 
But    Z  5CX=  Z  A  +Z  B  (?)  (108). 
/.Z  4  +  Z£  +  Z^C'l*  =  2  rt.  ^  =  180°    (Ax.  6).     Q.E.D. 


BOOK  I  43 

111.  COR.   The  sum  of  any  two  angles  of  a  triangle  is  less  than 
two  right  angles.    (See  Ax.  5.) 

112.  COR.   A  triangle  cannot  have   more  than  one  right  angle  01 
more  than  one  obtuse  angle. 

113.  COR.   Two  angles  of  every  triangle  are  acute.     (See  112.) 

114.  THEOREM.   The  acute  angles  of  a  right  triangle  are  comple- 
mentary. 

Proof :    Their  sum  =  1  rt.  Z    (110  and  Ax.  2).       Hence 

they  are  complementary.     (See  20.) 

115.  COR.    Each  angle  of  an  equiangular  triangle  is  60°. 

116.  THEOREM.    If  two  right  triangles  have  an  acute  angle  of  one 
equal  to  an  acute  angle  of  the  other,  the  remaining  acute   angles 
are  equal.     (See  114  and  48.) 

117.  THEOREM.    If  two  triangles  have  two  angles  of  the  one  equal 
to  two  angles  of  the  other,  the  third  angle  of  the  first  is  equal  to  the 
third  angle  of  the  second.     (See  110  and  Ax.  2.) 

118.  THEOREM.    Two  triangles  are  equal  if  a  side  and  any  two 
angles  of  the  one  are  equal  respectively  to  a  homologous  side  and  the 
two  homologous  angles  of  the  other. 

Proof  :    The  third  Z  of  one  A  =  third  Z  of  other  A  (117). 
.-.  the  A  are  =    (54). 

119.  THEOREM.   Two  right  triangles  are  equal  if  a  leg  and  the  op- 
posite acute  angle  of  one  are  equal  respectively  to  a  leg  and  the  opposite 
acute  angle  of  the  other.     (See  118.) 


Ex.   1.    In  the  figure  of  107,  if  Z  A  =  40°  andZ  C  =  70°,  how  many 
degrees  are  there  in  Z  ABXt  in  Z  ABC  ? 

Ex.   2.   If  each  of  the  equal  angles  of  an  isosceles  triangle  is  50°,  how 
many  degrees  are  there  in  the  third  angle  ? 

Ex.   3.   If  one  of  the  acute  angles  of  a  right  triangle  is  25°,  how  many 
degrees  are  there  in  the  other? 

Ex.  4.   State  and  prove  the  converse  of  114. 


44 


PLANE   GEOMETRY 


120.  THEOREM.   If  two  angles  of  a  triangle  are  equal,  the  triangle  is 
isosceles.     [Converse  of  55.]  8 

Given:  A  ABC;  Z^i=ZC. 
To  Prove  :  AB  —  BC. 
Proof:  Suppose  BX  drawn 
_L  to  AC. 

In  rt.  A  ABX  and  CBX,  BX 


.'.  A  ABX=  A  CBX  (?)    (119).     .'.  AB  =  BC  (?).          Q.E.D. 

121.  THEOREM.   An  equiangular  triangle  is  equilateral. 

122.  THEOREM.   If  two  sides  of  a  triangle  are  unequal,  the  angle 
opposite  the  greater   side  is  greater  than   the   angle   opposite  the 
less  side. 

Given  :    A  ABC ;   AB  >  AC. 

To  Prove  :    Z  ACB  >  Z  B. 

Proof  :    On  AB  take  AE  =  AC. 
[We  may,  because  AB  >  AC.~\ 

Draw  CE  and  let  Z  AEC=  x.   c  B 

Z  AEC  is  an  ext.  Z  of  A  C3S  (?).     .-.  Z  z  >  Z  B  (109). 

Also,  Z  4  OR  =  Z  ^LE<7  =  Z  x  (?)  (55).     Again,  Z  ACB  > 
Z  z  (?)  (Ax.  5).     .-.  Z  4CB  >  Z.B  (Ax.  11).  Q.E.D. 

123.  THEOREM.   If  two  angles  of  a  triangle  are  unequal,  the  side 
opposite  the  greater   angle  is   greater  than  the  side  opposite  the 
less  angle. 

Given:  AABC\  /_ACB>/-B. 
To  Prove:    AB  >  AC. 
Proof :    In  Z  ACB,   suppose 
Z  BCE  constructed  =  Z  B. 
Then,  CE=  BE  (?)  (120). 
Also  AE  +  CE  >  AC  (?). 
/.  ,4E  +  BE  >  ^LC  (Ax.  6).     That  is,  AB  >  AC.  Q.E.D. 

124.  THEOREM.   The  hypotenuse  is  the  longest  side  of  a  right  tri- 
angle.    (See  123.) 


BOOK  I 


QUADRILATERALS 


45 


125.  A  quadrilateral  is  a  portion  of   a    plane  bounded 
by  four  straight  lines.      These    four   lines   are   called    the 
sides.     The   vertices  of  a  quadrilateral  are  the  four  points 
at  which  the  sides  intersect.     The  angles  of  a  quadrilateral 
are  the  four  angles  at  the  four  vertices.      The  diagonal  of 
a  rectilinear  figure  is  a  line   joining   two   vertices,  not  in 
the  same  side. 

126.  A  trapezium  is  a  quadrilateral  having  no  two  sides 
parallel. 

A  trapezoid  is  a  quadrilateral  -having  two  and  only  two 
sides  parallel. 

A  parallelogram  is  a  quadrilateral  having  its  opposite 
sides  parallel  (O). 


TRAPEZOID 


PARALLELOGRAM 
RHOMBOID 


SQUARE 


RECTANGLE 


127.  A   rectangle  is   a   parallelogram   whose   angles   are 
right  angles. 

A  rhomboid  is  a  parallelogram  whose  angles  are  not  right 

angles. 

128.  A  square  is  an  equilateral  rectangle. 

A  rhombus  is  an  equilateral  rhomboid. 
i 

129.  The  side  upon  which  a  figure   appears  to  stand  is 
called  its  base.     A  trapezoid  and  all  kinds  of  parallelograms 
are  said  to  have  two  bases, — the  actual  base  and  the  side 
parallel  to  it.    The  non-parallel  sides  of  a  trapezoid  are  some- 
times called  the  legs.     An  isosceles  trapezoid  is  a  trapezoid 


46  PLANE   GEOMETRY 

whose  legs  are  equal.  The  median  of  a  trapezoid  is  the 
line  connecting  the  midpoints  of  the  legs.  The  altitude 
of  a  trapezoid  and  of  all  kinds  of  parallelograms  is  the 
perpendicular  distance  between  the  bases. 

130.   THEOREM.    The  opposite  sides  of  a  parallelogram  are  equal. 
Given  :   O  LMOP.  L 

To  Prove  :    LM  =  PO   and 
LP  =  MO. 

Proof  :    Draw  diagonal  PM. 
In  A  LMP  and  OMP,  PM  = 


.'.  ALMP  =  A  OMP  (?)    (54). 

.'.  LM  =  PO  and  LP  =MO  (?)    (27).  Q.E.D. 

131.  COR.   Parallel  lines  included  between  parallel  lines  are  equal. 
(See  130.) 

132.  COR     The  diagonal  of  a  parallelogram  divides  it  into  two 
equal  triangles. 

133.  COR.  The  opposite  angles  of  a  parallelogram  are  equal.  (See  27.) 

134.  THEOREM.    If  the  opposite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelogram.     [Converse  of  130.] 

Given:  Quadrilateral  ABCD-9  A  B 

AB  =  DC  l   AD  =  BC. 

To  Prove  :    A  BCD  is  a  O. 

Proof  :    Draw  diagonal  BD.        /        ...•••""" 
In  &ABD    and     CBD,    BD  = 
BD  (?)  ;     AB  =  DC  (?),    and    tf 
AD  =  BC  (?).      .'.  A  ABD  =  A  CBD   (?)    (58). 

Hence  Z  a  =  Z  i  (?).     Therefore  AB  is  ||  to  DC  (?)  (101). 

Also,  Z  y  =  Zz    (?).     Therefore  ^D  is  ||  to  .BC  (?). 

Hence  ABCD  is  a  parallelogram  (Def.  126).  Q.E.D. 


BOOK  I  47 

135.  THEOREM.   If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Given :    Quadrilateral  A  BCD  ;  AB  =  CD  and  AB  \\  to  CD. 
To  Prove:   A  BCD  is  a  O. 

Proof :    Draw  diagonal  BD.       In  A  ABD  and  CDD,  DD  = 
DD(?);  ^D  =  CD  (?);  and  Za  =  Z.i  (?)  (97). 
.-.  A.1DD  =  ACDD  (?)   (52). 

Hence  /_y=/.x    (?j.     .-.   ^4D    is  ||  to  DC  (?)  (101). 
.'.ABCD  is  a  parallelogram  (?)   (126).  Q.E.D. 

136.  COR.  Any  pair  of  adjoining  angles  of  a  parallelogram  are  sup- 
plementary.   (See  100.) 

137.  THEOREM.  The  diagonals  of  a  parallelogram  bisect  each  other. 

Given  :    O  EFGH  ;    diago-  F — 

nals  EG  and  FH  intersecting 
at  X. 

To  Prove :    FX  =  XH   and 

GX  —  XE. 


Proof  :  In  A  FXG  and  EXH,     E 

FG  =  ^JT  (?)   (130)  ;  Z  a  =  Z  0  and  Z  c  =  Z  /•  (?)   (97). 
.-.  A  FXG  =  A  EXH  (?)  (54). 
/.  FX  =  XH  and  GX  =  XE  (?)   (27).  Q.B.D. 


138.  THEOREM.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  figure  is  a  parallelogram. 

Given:    (?).       To   Prove:    (?).       Proof:    In  A  FXG  and 

EXH  show  three  parts  of  one  =  etc.     Hence  certain  A  are 
=  (?).     Then  two  lines  are  II  (?).     Also  =  (?).    Now  use  135. 


Ex.  1.   In  the  figure  of  137,  if  Z  a  =  20°  and  Z  c  =  30°,  find  the  four 
angles  at  X. 

Ex.  2.  If  one  angle  of  a  parallelogram  is  65°,  find  the  other  three.     If 
one  is  90°,  find  the  others. 

Ex.  3.   State  and  prove  the  converse  of  136. 


48  PLANE  GEOMETRY 

139.  THEOREM.  Two  parallelograms  are  equal  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and  the  in- 
cluded angle  of  the  other. 

c          M N 


A  D 

Given  :    UJAC  and  LN;  AB  =  LM;   AD  =  LO  ;  Z  A  =  /_  L. 
To  Prove :   The  UJ  are  = . 

Proof:    Superpose    UJ  ABCD  upon  n  LMNO,  so  the    equal 
angles  A  and  L  coincide,  AD  falling  along  LO  and  AB  along  LM. 
Point  D  will  coincide  with  point  o  [AD  —  LO  (Hyp.)]. 
Point  B  will  coincide  with  point  M  [AB  —  LM  (Hyp.)]. 
BC  and  MN   are  both  ||  to   LO  (?)  (126). 
.-.  BC  falls  along  MN  (?)  (92). 
CD   and   JVO   are    both  ||  to   LM   (?). 
.-.CD  falls  along  JVO  (?). 
Hence  C  will  fall  exactly  upon  N  (38). 
.*.  the  figures  coincide,  and  are  equal  (?)  (28).       Q.B.D. 

140.  THEOREM.  Two  rectangles  are  equal  if  the  base  and  altitude  of 
one  are  equal  respectively  to  the  base  and  altitude  of  the  other.  (See  139.) 

141.  THEOREM.     The  diagonals  of  a  rhombus  (or  of  a  square)  are 
perpendicular  to  each  other,  bisect  each  other,  and  bisect  the  angles  of 
the  rhombus  (or  of  the  square).  o 

Given :   Rhombus  ABCD  ;   di- 
agonals AC  and  BD. 

To  Prove:  AC  J_  to  BD;  AC 

and  BD  bisect  each  other ;    and 

they  bisect  A  DAB,  ABC,  etc.      A  B 

Proof  :  Point  A  is  equally  distant  from  B  and  D  (?)   (128). 
Point  C  is  equally  distant  from  B  and  D  (?). 
.'.  AC  is  _L  to  BD  (?)   (70).  Q.E.D. 


BOOK  I 


Also  AC  and  BD  bisect  each  other  (?)  (137).  Q.E.D. 

The  A  at  A  are  =  (?)  (66,  II).     Etc.  Q.E.D. 

The  proof  if  the  figure  is  a  square  is  exactly  the  same. 

142.  THEOREM.    The  line  joining  the  midpoints  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side  and  equal  to  half  of  it. 

Given:  A  ABC',  M,  the 
midpoint  of  AB\  P,  the 
midpoint  of  BC>  line  MP. 

To  Prove :  MP  II  to  AC 
and  MP  =  1  AC. 

Proof:  Suppose  AE  is 
drawn  through  A,  II  to  BC  A 

and  meeting  MP  produced  at  E.  In  A  AEM  and  BPM,  AM 
=  BM  (Hyp.);  ^x  =  Ze  (?)  and  Z  o  =  Z  B  (?)  (97). 
.'.  A  AEM  =  A  BPM  (?)  (54). 

Hence,  AR  =  BP  (?).     But  BP  =  PC  (?).    /.  AE  =  PC  (?). 

.-.  ACPE  is  a  O  (?)  (135). 

Hence  EP  or  JfP  is  II  to  AC  (?).  Q.E.D. 

Also,  EP  =  AC  (?)  (130).    But  MP  =  E3f  (?)  (27). 

.-.  MP=\  RP  =  \  AC  (Ax.  6).  Q.E.D. 

143.  THEOREM.     The   line  bisecting  one  side  of  a  triangle  and 
parallel  to  a  second  side,  bisects  also  the  third  side. 

Given :  A  ABC ;  MP  bisecting 
AB  and  II  to  AC. 

To  Prove :  MP  bisects  BC  also. 

Proof :  Suppose  MX  is  drawn 
from  jf,  the  midpoint  of  AB  to 
X,  the  midpoint  of  BC. 

MX  is  II  to  AC  (142)  ;    but  MP    A  ~~O 

is  II  to  AC  (Hyp.). 

.*.  MX  and  MP  coincide  (?). 

That  is,  MP  bisects  BC.  Q.E.D. 


50  PLANE   GEOMETRY 

144.  THEOREM.     The   line   bisecting  one  leg  of  a  trapezoid  and 
parallel  to  the  base  bisects  the  other  leg,  is  the  median,  and  is  equal 
to  half  the  sum  of  the  bases. 

Given :  Trapezoid  ABCD ;  M,  the  midpoint  of  AB ;  MP  II  to 
AD,  meeting  CD  at  P. 

To  Prove:    I.    P  is  the 
midpoint  of  CD. 

II.  MP  is  the  median.  MZ. ^3 \P 

III.  MP  =  |  (AD  +  BC). 

Proof:  I.  Draw  diagonal 
BD,  meeting  MP  at  R. 

MP  is  II  to  BC  (94).    In  A  ABD,  MR  bisects  BD  (143). 
In  ABDC,  RP  bisects  CD  (?)  (143). 
That  is,  P  is  the  midpoint  of  CD. 
II.    MP  is  the  median  (Def.  129). 
III.    MR  =  1  AD  (142)  and  RP  =  ^  BC  (?). 
.'.  MP  =  1  (AD  +  BC)  (Ax.  2).  Q.E.D. 

145.  THEOREM.     The  angles  adjoining  each  base  of  an  isosceles 
trapezoid  are  equal. 

Given:    Trapezoid   AC;  AB  =  CD. 

To  Prove :  Z.  A  =  Z.  D  and 
Z.  ABC  =  Z  (7. 

Proof :  Suppose  BXis  drawn 
through  B  and  II  to  CD.  BX  = 
CD  (130);  AB  =  CD  (Hyp.).  A~~  "T> 

/.  AB  =  BX  (?),  and  Z.  A  =  Z  a  (?),  and  Z.  a  =  Z.  D  (?)  (98). 

.'.  Z.  A  =  Z.  D  (Ax.  1). 

Again,  Z  C  is  supp.  of  Z  D  (?).   Etc.  Q.E.D. 

146.  THEOREM.     If  the  angles  at  the  base  of  a  trapezoid  are  equal, 
the  trapezoid  is  isosceles. 

Given:  (?).     To  Prove:  (?). 

Proof:    Suppose  BX    is  drawn  II  to  CD.     Z  a  =  Z  D  (?); 
Z.A  =  Z.D   (?)      .-.  Z^  =  Z  a  (?\     .\AB  =  BX(?}.  Etc. 


BOOK  I 


NOTE.  The  verb  "to  intersect"  means 
merely  "  to  cut."  In  geometry,  the  verb  "  to 
intercept "  means  "  to  include  between"  Thus 
the  statement "  AB  and  CD  intercept  XY  on 
the  line  EF"  really  means,  "AB  and  CD  A 
intersect 'EF  and  include  XY,  a  part  of  EF, 
between  them." 


51 


7     i 


L 


147.  THEOREM.     Parallels  intercepting  equal  parts  on  one  trans- 
versal intercept  equal  parts  on  any  transversal. 

Given  :   Us  AB,  CD,  EF.  GH,  IJ  intercepting  equal  parts  -4C, 
CEt  EG,  Gl,  on  the  transversal  A/ 

AI,  and  cutting  transversal  BJ. 

To  Prove  :  BD  =  DF  =  FH 
—HJ. 

Proof:  The  figure  ABFE  is 
atrapezoid  (?).  CD  bisects  AE 
and  is  II  to  EF  (Hyp.). 

.*.  D  is  midpoint  of  BF  (?). 
That  is,  BD  —  DF. 

Similarly,  CDHG  is  a  trapezoid  and   EF  bisects  DH  (?). 
That  is,  DF=  FH. 

Similarly,  FH  =  HJ.    .\BD  =  DF=FH  =  HJ  (Ax.  1).  Q.E.D. 

148.  THEOREM.     The  midpoint  of  the  hypotenuse  of  a  right  triangle 
is  equally  distant  from  the  three  vertices. 

Given:  Rt.  A  4.8(7;  Jf,  the  midpoint  of 
hypotenuse  AB. 

To  Prove  :  AM  =  CM 


Proof:  Suppose  MX  is  drawn  II  to  BC, 
meeting  AC  at  X.  X  is  midpoint  of 
AC  (?)  (143).  MX  is  JL  to  AC  (95). 

That  is,  MX  is  J_  to  AC  at  its  midpoint, 
SUL&AM=MC  (?)  (67). 

But    AM=BM    (Hyp.). 

.'.  AM  =  CM  =  BM   (Ax.  1).  Q.E.D. 


52 


PLANE   GEOMETRY 


149.  THEOREM.     The  median  of  a  trapezoid  is  parallel  to  the  bases 
and  equal  to  half  their  sum. 

[This  is  another  form  of  stating  the  theorem  of  144.] 

150.  THEOREM.  The  perpendiculars  from  the  vertices  of  a  triangle 
to  the  opposite  sides  meet  in  a  point. 


A'-. 


f 


Given :    A  ABC,  AX _L  to  BC,  BY  _L  to  AC,  and  CZ  _L  to  AB. 
To  Prove :     These  three  Js  meet  in  a  point. 

Proof  :  Through  A  suppose  RS  drawn  II  to  BC  ;  through  JS, 
T8  II  to  AC ;  through  C,  ET  II  to  AB,  forming  A  EST. 

The  figure  ABCE  is  a  O  (Const.)  and  ABTC  is  a  O  (?). 
/.  EC  =  AB  and  CT  =  AB  (?)  (130).  .-.  EC  =  CT  (Ax.  1). 

Now  CZ  is  ±  to  ET  (?)  (95). 

That  is,  CZ  is  _L  to  .RT  at  its  midpoint. 

Similarly  AX  is  _L  to  .RS  at  its  midpoint. 

And  BY  is  _L  to  TS  at  its  midpoint. 

Therefore  AX,  BY,  CZ  meet  at  a  point  (?)  (85).      Q.E.D. 


Ex.  1.  Draw  the  three  altitudes  of  an  obtuse  triangle  and  prolong 
them  until  they  meet. 

Ex.  2.  Prove  that  each  of  the  three  outer  triangles  in  the  figure  of 
150  is  equal  to  A  AB C. 

Ex.  3.  Prove  that  any  altitude  of  A  EST  is  double  the  parallel  alti- 
tude of  A  ABC.  [Use  143  and  130.] 


BOOK   I 


53 


151.  THEOREM.  The  point  at  which  two  medians  of  a  triangle  inter- 
sect is  two  thirds  the  distance  from  either  vertex  of  the  triangle  to  the 
midpoint  of  the  opposite  side. 

Given  :  A  ABC,  BD  and   CE  two  medians  intersecting  at  O. 

To  Prove  :   BO  =  f  BD  and  CO  =  f  CE. 

Proof  :  Suppose  H  is  the  midpoint  of  BO  and  J  is  the  mid- 
point of  CO.     Draw  ED,  DI, 
IH,  HE. 

In  A  ABC,  DE  is  II  to  BC 
and  =  J  BC  (?)  (142). 

In   A  OBC,  HI  is  II  to  BC 
and  =  JBC(?). 

.-.  ED  =  HI  (Ax.  1),  and 
ED  is  II  to  HI  (?)  (94). 


_ 
B  c 

/.  HO  =  OD  and  10  =OE  (?)  (137). 
.'.  BH—  HO  =  OD  and  CI=  IO  =  OE  (Ax.  1). 
That  is,  BO  —  2  •  OD  =  f  BD,  and  CO  =  2  •  #O  =  f  CE.   Q.E.D. 

152.  THEOREM.  The  three  medians  of  a  triangle  meet  in  a  point 
which  is  two  thirds  the  distance  from  any  vertex  to  the  midpoint  of  the 
opposite  side. 

Proof  :  Suppose  AX  is  the  third  median  of  A  ABC  and 
meets  BD  at  O1. 

Then  BOr  =  f  BD  and  BO  =  f  BD  (?)  (151)  . 

.-.  BOf  =  BO  (?).  That  is  o'  coincides  with  O  and  the 
three  medians  meet  at  O  which  is  the  distance,  etc.  Q.E.D. 


Ex.  1.    In  the  figure  of  151,  prove  EH  =  \A  0  =  DI,  by  142. 

Ex.  2.  In  the  figure  of  151,  if  BO  AC,  prove  the  angle  BEG  obtuse. 
[Use  87.] 

Ex.  3.  If  one  angle  of  a  rhombus  is  30°,  find  all  the  angles  of  the 
four  triangles  formed  by  drawing  the  diagonals. 

Ex.  4.  Show  that  any  trapezoid  can  be  divided  into  a  parallelogram 
and  a  triangle  by  drawing  one  line. 

Ex.  5.  Prove  that  every  right  triangle  can  be  divided  into  two  isos- 
celes triangles  by  drawing  one  line.  [Use  148.] 


54  PLANE   GEOMETRY 


POLYGONS 

153.  A  polygon    is   a  portion    of    a    plane    bounded    by 
straight  lines.     The  lines  are  called  the  sides.     The  points 
of  intersection  of  the  sides   are   the  vertices.     The   angles 
of  a  polygon  are  the  angles  at  the  vertices. 

154.  The  number  of  sides  of  a  polygon  is  the  same  as  the 
number  of  its  vertices  or  the  number  of  its  angles.      An 
exterior  angle  of  a  polygon  is  an  angle  without  the  polygon, 
between  one  side  of  the  polygon  and  another  side  prolonged. 

155.  An  equilateral   polygon  has   all  of   its   sides   equal 
to   one   another.     An   equiangular   polygon   has    all   of  its 
angles  equal  to  one  another. 

156.  A  convex  polygon  is  a  polygon  no  side  of  which  if 
produced  will  enter  the   surface  bounded  by  the  sides  of 
the  polygon.     A  concave  polygon  is  a  polygon  two  sides  of 
which   if  produced  will  enter  the  polygon. 


EQUILATERAL  EQUIANGULAR  CONCAVE,   OB 

CONVEX  POLYGONS  RE-ENTRANT 

NOTE.  A  polygon  may  be  equilateral  and  not  be  equiangular  ;  or 
it  may  be  equiangular  and  not  be  equilateral.  The  word  "  polygon  " 
is  usually  employed  to  signify  convex  figures. 

157.  Two  polygons  are  mutually  equiangular  if  for  every 
angle  of  the  one  there  is  an  equal  angle  in  the  other  and 
similarly  placed.  Two  polygons  are  mutually  equilateral, 
if  for  every  side  of  the  one  there  is  an  equal  side  in  the 
other,  and  similarly  placed. 


BOOK  I  55 


158.  Homologous    angles    in    two    mutually    equiangular 
polygons   are  the  pairs  of  equal  angles.     Homologous  sides 
in  two  polygons  are  the  sides  between  two  pairs  of  homolo- 
gous angles. 

159.  Two  polygons  are  equal  if  they  are  mutually  equi- 
angular and  their  homologous  sides  are  equal  ;    or  if  they 
are  composed  of  triangles,  equal  each  to  each  and  similarly 
placed.     (Because  in  either  case  the  polygons  can  be  made 
to  coincide.) 

160.  Two   polygons   may  be   mutually  equiangular  without  being 
mutually  equilateral ;  also,  they  may  be  mutually  equilateral  without 
being  mutually  equiangular  —  except  in  the  case  of  triangles. 


The  first  two  figures  are  mutually  equilateral  but  not  mutually  equi- 
angular. The  last  two  figures  are  mutually  equiangular  but  not  mutu- 
ally equilateral. 

161.    A  3-sided  polygon  is  a  triangle. 

A  4-sided  polygon  is  a  quadrilateral. 
A  5-sided  polygon  is  a  pentagon. 
A  6-sided  polygon  is  a  hexagon. 
A  7-sided  polygon  is  a  heptagon. 
..  An  8-sided  polygon  is  an  octagon. 
A  10-sided  polygon  is  a  decagon. 
A  12-sided  polygon  is  a  dodecagon. 
A  15-sided  polygon  is  a  pentedecagon. 
An  n-sided  polygon  is  called  an   n-gon. 

Ex.  Draw  a  pentagon  and  all  the  possible  diagonals  from  one  vertex. 
How  many  triangles  are  formed?  Draw  a  decagon  and  the  diagonals 
from  one  vertex.  How  many  triangles  are  thus  formed  ?  Construct  a  20- 
gon  and  the  diagonals  from  one  vertex.  How  many  triangles  are  formed  ? 


56 


PLANE   GEOMETRY 


162.  THEOREM.   The  sum  of  the  interior  angles  of  an  n-gon  is  equal 
to  (n-2)  times  180°. 

Given  :  A  polygon  having 
n  sides. 

To  Prove  :  The  sum  of  its 
interior  A  =  (n  -  2)  .  180°. 

Proof :  By  drawing  all  pos- 
sible diagonals  from  any  vertex 
it  is  evident  that  there  will 
be  formed  (n  —  2)  triangles. 
The  sum  of  the  A  of  one  A  =  180°  (?)  (110). 

/.  the  sum  of  the  A  of  (n-2)  A  =  (n-2)  180°  (Ax.  3). 

But  the  sum  of  the  A  of  the  triangles  =  the  sum  of  the  A 
of  the  polygon  (Ax.  4). 

.-.  sum  of  A  of  the  polygons  (w— 2)  180°  (Ax.  1).    Q.E.D. 

163.  COR.   The  sum  of  the  interior  angles  of  an  n-gon  is  equal  to 
iSo°n  -  360°. 

164.  COR.   Each  angle  of  an  equiangular  n-gon  =  \n~ 2  )  *  °  . 

'n/ 

165.  COR.   The  sum  of  the  angles  of  any  quadrilateral  is  equal  to 
four  right  angles. 

166.  COR.   If  three  angles  of  a  quadrilateral  are  right  angles,  the 
figure  is  a  rectangle. 


Ex.  1.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
pentagon  ?  of  an  equiangular  pentedecagon  ?  of  a  30-gon  ? 

Ex.  2.  If  two  angles  of  a  quadrilateral  are  right  angles,  what  is  true  of 
the  other  two  ? 

Ex.  3.  How  many  sides  has  that  polygon  the  sum  of  whose  interior 
angles  is  equal  to  20  rt.  A  ? 

Ex.  4.  How  many  sides  has  that  equiangular  polygon  each  of  whose 
angles  contains  160°? 

Ex.  5.  If  in  the  figure  of  105,  Za  =  65°,  how  many  degrees  are  there 
in  each  of  the  other  angles  of  the  figure  ? 


BOOK  1 


57 


167.  THEOREM.  If  the  sides  of  a  polygon  be  produced,  in  order, 
one  at  each  vertex,  the  sum  of  the  exterior  angles  of  the  polygon  will 
equal  four  right  angles,  that  is,  360°. 

Given  :    A  polygon  with  sides 
prolonged  in  succession  forming 
the  several  exterior  angles  «, 
<?,  c?,  etc. 

To  Prove:    Za  +  Z6  + 

Z  d  +  etc.  =  4  rt.  A  =  360°. 

Proof  :    Suppose  at  any  point 
in  the  plane,  lines  are  drawn 
parallel  to  the  several  sides  of  the  given  polygon,  extending 
in  the  same  direction,  and  forming  angles  J.,  £,  C,  D,  etc. 

Then  Z  A  +  Z.  B  +  /.  C  +  Z  D  +  Z  E      \         / 
+  Z  F+Z(?  =  4  rt.  A  (?)   (47). 

But  Za  =  Z.A,  Zb  =  Z.B,  Z  <?  =  Z 
Zd=Z  A  etc.   (?)   (105). 


•     •   •"*  -'A 

DV.-:::..A.... 
£'/'{  G 


=  4  rt.  Zs  =  360°   (?)    (Ax.  6). 


Q.E.D. 


168.    COR.    Each  exterior  angle  of  an  equiangular  polygon  is  equal 


to 


4rt. 


that  is, 


^360° 


n 


169.   COR.    The  sum  of  the  exterior  angles  of  a  polygon  is  indepen- 
dent of  the  number  of  its  sides. 


Ex.  1.  How  many  degrees  are  there  in  each  exterior  angle  of  an 
equiangular  dodecagon  ?  of  an  equiangular  40-gon  ? 

Ex.  2.  If  any  angle  of  an  isosceles  triangle  is  60°,  the  triangle  is 
equiangular. 

Ex.  3.  Prove  the  theorem  of  162  by  drawing  lines  from  any  point 
within  the  triangle  to  all  the  vertices. 

Ex.  4.   State  the  theorems  which  deal  with  concurrent  lines. 

Concurrent  lines  are  lines  that  meet  in  a  common  point. 

Ex.  5.  How  many  sides  has  that  polygon  the  sum  of  whose  interior 
angles  exceeds  the  sum  of  the  exterior  angles  by  720°? 


PLANE  GEOMETRY 


SYMMETRY 

170.  A  figure  is  symmetrical  with  respect  to  a  line  if,  by 
using  that  line  as  an  axis,  the  part  of  the  figure  on  one  side 
of  the  line  may  be  folded  over,  and  will  exactly  coincide  with 
the  part  on  the  other  side.    This  line  is  an  axis  of  symmetry. 

171.  A  figure  is  symmetrical  with  respect  to  a  point  if  this 
point  bisects  every  line  drawn  through  it  and  terminated 
(both  ways)  in  the  boundary  of  the  figure. 

This  point  is  the  center  of  symmetry. 

172.  It  is  evident  that  the  axis  of  symmetry  bisects  at 
right  angles  every  line  joining  two  symmetrical  points ;  and 
that  the  center  of  symmetry  bisects  every  line  joining  any 
pair  of  points  symmetrical  with  respect  to  it. 


Examples  of  symmetry  are  given  in  these  figures. 
First  figure  is  symmetrical  with  respect  to  XX'  as  an  axis.     (Why?) 
Second  figure  is  symmetrical  with  respect  to  0  as  a  center.     (Why?) 
P'  and  /Y  are  symmetrical  with  respect  to  XX'  as  an  axis.     (Why?) 
A  and  A't  B  and  B'  etc.  are  symmetrical  with  respect  to  0  as  a  center. 
(Why?)     XX'  is  ±  to  PfPlf  and  bisects  it.    A  0  =  A'O,  BO  =  B'O,  etc. 

173.  In  order  to  prove  that  a  line  is  an  axis  of  symmetry 
it  is  necessary  to  show  that  it  satisfies  the  condition  of  170. 

In  order  to  prove  that  a  point  is  a  center  of  symmetry  it  is 
necessary  to  show  that  it  satisfies  the  condition  of  171. 


BOOK   I 


59 


174.  THEOREM.   If  two  lines  are  symmetrical  with  respect  to  a 
center,  they  are  equal  and  parallel. 

Given  :  AB  and  ES  symmet- 
rical with  respect  to  o,  that  is, 
every  line  through  O,  termi- 
nated in  AB  and  ES  is  bisected 
ftt  O;  A3  and  BE,  two  such 
lines. 

To  Prove :    AB  =  ES  and  AB  II  to  ES. 

Proof  :    Draw  AE  and  BS.     AO=  OS  and  BO  =  OE  (Hyp.). 

.-.   ABSE  is  a  O(?)  (138). 

.'.  AB  =  ES  (?)  and  AB  is  11  to  ES  (?).  Q.E.D. 

175.  THEOREM.   If  a  diagonal  of  a  quadrilateral  bisects  two  of  its 
angles,  this  diagonal  is  an  axis  of  symmetry. 

Given:   Quadrilateral  B 

ABCD;  AC  a  diagonal  bisect- 
ing Z  BAD  and  Z  BCD. 

To  Prove  :  AB  CD  symmet- 
rical with  respect  to  AC. 

Proof :  In  A  ABC  and 
ADC,  AC  =  AC  (?). 

Z  BAG  =  Z.  DAC   (?)  and  o 


Hence  A  ABC=  A  ^IDC  (?). 

.*.  AC  is  an  axis  of  symmetry  (?).  Q.E.D. 

176.  COR.    The  diagonal  of  a  square  or  of  a  rhombus  is  an  axis 
of  symmetry.     (Why  ?) 

177.  COR.    The  diagonal  of  a  rectangle  or  of  a  rhomboid  is  not  an 
axis  of  symmetry.      (Why  not?) 

Ex.  1.  Is  the  altitude  of  an  equilateral  triangle  an  axis  of  symmetry? 

Ex.  2.  Is  the  altitude  of  an  isosceles  triangle  an  axis  of  symmetry? 

Ex.  3.  Are  all  altitudes  of  all  triangles  axes  of  symmetry? 

Ex.  4.  Has  an  isosceles  trapezoid  an  axis  of  symmetry? 


60 


PLANE   GEOMETRY 


178.  THEOREM.   If  a  figure  is  symmetrical  with  respect  to  two  per- 
pendicular axes,  it  is  symmetrical  with  respect  to  their  intersection  as  a 
center. 

Given :  Figure  MN  sym- 
metrical with  respect  to  the 
J_  axes  XXf  and  YYr  which 
intersect  at  O. 

To  Prove :  Figure  MN  is 
symmetrical  with  respect  to 
O  as  a  center. 

Proof:  Take  any  point  P 
in  the  boundary.  Draw  PB 
-L  to  FF',  intersecting  TTr  at  A  and  meeting  the  boundary 
at  B.  Draw  BR  _L  to  XX1,  intersecting  XX1  at  G  and  meet- 
ing the  boundary  at  E.  Draw  AC,  OP,  OR. 

[The  demonstration  is  accomplished    by  proving  BOP  a 
straight  line,  bisected  at  O.] 

PB  is  II  to  XX'  and  BR  is  II  to  FF'  (?)  (93). 

Hence  ABCO  is  a  O  (?). 

.'.BC=AO  (?).      But  J?C=CB(?)  (172). 

.'.  AO=CR  (Ax.  1). 

.*.  ACRO  is  a  O  (?)  (135).     .-.  RO  is  =  and  II  to  AC  (?). 

Similarly  CO  is  =  and  II  to  AP ;  hence  ACOP  is  a  E7  (?). 

.-.PO  is  =  and  II  to  AC  (?). 

/.  POR  is  a  straight  line  (?)  (92)  and  PO  =  OR  (?). 

But  P  is  any  point,  so  PO.R  is  any  line  through  o. 

Hence  O  is  a  center  of  symmetry  (171).  Q.E.D. 

LOCUS 

179.  The  locus  of  a  point  is  the  series  of  positions  the  point 
must  occupy  in  order  that  it  may  satisfy  a  given  condition. 
It  is  the  path  of  a  point  whose  positions  are  limited  or  denned 
by  a  given  condition,  or  given  conditions. 

180.  Explanatory.    I.    If  a  point  is  moving  so  that  it  is 
always   one  inch  from  a  given  indefinite  straight  line,  the 


BOOK  1  61 


point  may  occupy  any  position  in  either  of  two  indefinite  lines, 
one  inch  from  the  given  line,  parallel  to  it,  and  one  on  either 
side  of  it.  And  this  point  cannot  occupy  any  position  which 
is  not  in  these  lines.  Hence,  the  locus  of  points  at  a  given 
distance  from  a  given  line  is  a  pair  of  parallels  to  the  given 
line,  one  on  either  side  of  it,  and  at  the  given  distance  from  it. 

II.  If  a  point  is  moving  so  that  it  is  always  equally  distant 
from  two  parallels,  it  must  move  in  a  third  parallel  midway 
between  them.     Hence,  the  locus  of  points  equally  distant 
from  two  parallels  is  a  third  parallel  midway  between  them. 

III.  The  method  of  proving  that  a  certain  line  or  group 
of  lines  is  the  locus  of  points  satisfying  a  given  condition, 
consists  in  proving  that  every  point  in  the  line  fulfills  the 
given   requirement,  and   that  there  is  no  other  point  that 
fulfills  it.     In  the  above  illustrations  it  is  evident  that  every 
point  in  the  lines  which  were  called  the  "locus,"  did  fulfill 
the  conditions  of  the  case.     It  is  also  evident  that  there  is 
no  point  outside   these    "loci"  which  does   so    fulfill    the 
conditions.     That   is,   these    "loci"    contain  all  the  points 
described  and  no  others. 

IV.  THEOREM.   The  locus  of  points  equally  distant  from  the  ex- 
tremities of  a  line  is  the  perpendicular  bisector  of  the  line. 

Proof:  Every  point  in  the  _L  bisector  of  a  line  is  equally 
distant  from  its  extremities  (67).  And  also,  there  is  no 
point  outside  the  _L  bisector  which  is  equally  distant  from 
the  extremities  of  a  line  (69).  Hence  it  is  the  locus  of 
points  equally  distant  from  the  extremities  of  the  line. 

V.  THEOREM.   The  locus  of  points  equally  distant  from  the  sides 
of  an  angle  is  the  bisector  of  the  angle. 

Proof :    Like  the  preceding  proof.      [Use  79,  80,  81.] 

VI.  The  locus  of  the  vertices  of  all  the  isosceles  triangles 
that  can  be  constructed  on  a  given  base  is  the  perpendicular 
bisector  of  the  base.     (Same  as  IV.) 


62  PLANE   GEOMETRY 


CONCERNING  ORIGINAL  EXERCISES 

181.  In  the  original  work  which  this  text  contains,  the 
pupil  is  expected  to  state  the  hypothesis  and  conclusion  of 
each  theorem,  and  apply  them  to  an  appropriate  figure.     He 
is   expected   to   state    completely  and  logically    the  proof, 
giving  a  correct  reason  for  every  declarative  statement. 

In  many  of  these  exercises,  suggestions  are  made  and  such 
assistance  is  given  as  experience  has  shown  average  pupils 
require.  This  is  done  in  order  that  the  learner  may  be  en- 
couraged toward  definite  accomplishment,  which  is  one  of 
the  greatest  incentives  to  further  effort. 

To  apply  the  knowledge  acquired  from  the  preceding  pages 
is  now  the  student's  task.  His  fascination  for  this  science 
will  depend  largely  upon  the  success  of  his  efforts  at  proving 
originals.  Therefore,  many  obstacles  will  be  removed  or  modi- 
fied, and  no  trouble  will  be  spared  in  making  the  mastery  of 
this  department  of  geometry  both  agreeable  and  profitable. 

The  student  should  not  draw  a  special  figure  for  a  general 
proposition.  That  is,  if  "triangle  *  is  specified, he  should  draw 
a  scalene  and  not  an  isosceles  or  a  right  triangle  ;  and  if 
"quadrilateral"  is  mentioned,  he  should  draw  a  trapezium 
and  not  a  parallelogram  or  a  square. 

SUMMARY.     GENERAL  DIRECTIONS  FOR  ATTACKING  EXERCISES 

182.  A  triangle  is  proved  isosceles  by  showing  that  it  contains  two 
equal  sides,  or  two  equal  angles. 

183.  A  triangle  is  proved  a  right  triangle  by  showing  that  one  of  its 
angles  is  a  right  angle,  or  two  of  its  angles  are  complementary,  or  one  of 
its  angles  is  equal  to  the  sum  of  the  other  two. 

184.  Right  triangles  are  proved  equal,  by  showing  that  they  have : 

(1)  Hypotenuse  and  acute  angle  of  one  =  etc. 

(2)  Hypotenuse  and  leg  of  one  =  etc. 

(3)  The  legs  of  one  =  etc. 

(4)  Leg  and  adjoining  angle  of  one  =  etc. 

(5)  Leg  and  opposite  angle  of  one  =  etc. 


BOOK  I  63 

185.  Oblique  triangles  are  proved  equal,  by  showing  that  they  have: 

(1)  Two  sides  and  the  included  angle  of  one  =  etc. 

(2)  One  side  and  the  adjoining  angles  of  one  =  etc. 

(3)  Three  sides   of  one  =  etc. 

186.  Angles  are  proved  equal,  by  showing  that  they  are : 

(1)  Equal  to  the  same  or  to  equal  angles. 

(2)  Halves  or  doubles  of  equals. 

(3)  Vertical  angles. 

(4)  Complements  or  supplements  of  equals. 

(5)  Homologous  parts  of  equal  figures. 

(6)  Base  angles  of  an  isosceles  triangle. 

(7)  Corresponding  angles,  alternate-interior  angles,  etc.  of  parallels. 

(8)  Angles  whose  sides  are  respectively  parallel  or  perpendicular. 

(9)  Third  angles  of  triangles  which  have  two  angles  of  one  =  etc. 

187.  Lines  are  proved  equal,  by  showing  that  they  are : 

(1)  Equal  to  the  same  or  to  equal  lines. 

(2)  Halves  or  doubles  of  equals. 

(3)  Distances  to  the  ends  of  a  line  from  any  point  in  its  perpendicu- 

lar bisector. 

(4)  Homologous  parts  of  equal  figures. 

(5)  Sides  of  an  isosceles  triangle. 

(6)  Distances  to  the  sides  of  an  angle  from  any  point  in  its  bisector. 

(7)  Opposite  sides  of  a  parallelogram. 

(8)  The  parts  of  one  diagonal  of  a  parallelogram  made  by  the  other. 

188.  Two  lines  are  proved  perpendicular,  by  showing  that  they  : 

(1)  Make  equal  adjacent  angles  with  each  other. 

(2)  Are  legs  of  a  right  triangle. 

(3)  Have  two  points  in  one,  each  equally  distant  from  the  ends  of  the 

other. 

189.  Two  lines  are  proved  parallel,  by  : 

(1)  The  customary  angle-relations  of  parallel  lines. 

(2)  Showing  that  they  are  opposite  sides  of  a  parallelogram. 

(3)  Showing  that  they  are  parallel  or  perpendicular  to  a  third  line. 

190.  Two  lines,   or  two  angles,  are  proved  unequal  by  the  usual 
axioms  and  theorems  pertaining  to  inequalities. 

[See  especially,  Ax.  5*;  Ax.  12;  6.8;  75;  76,  III;  77;  86  ;  87*;  90 
109*;  122*;  124.] 

*  These  refer  to  angles. 


64 


PLANE   GEOMETRY 


ORIGINAL   EXERCISES 

1.  A  line  cutting  the  equal  sides  of  an  isosceles  triangle  and  parallel 
to  the  base  forms  another  isosceles  triangle.     [Use  186  (7),  and  182.] 

2.  The  bisectors  of  the  equal  angles  of  an  isosceles  triangle  form,  witl. 
the  base,  another  isosceles  triangle.     [Use  186  (2).] 

B 

3.  If  the  exterior  angles  at  the  base  of  a  triangle 

are  equal,  the  triangle  is  isosceles.    [186  (4).] 

4.  If  from  any  point  in  the  base  of  an  isosceles  tri- 
angle a  line  be  drawn  parallel  to  one  of  the  equal 
sides  and  meeting  the  other  side,  an  isosceles  triangle     »• 
will  be  formed. 

To  Prove  :  A  DEC  isosceles. 


5.  If  the  median  of  a  triangle  is  perpendicular 
to  the  base,  the  triangle  is  isosceles.     [Use  187  (3).] 

6.  If  a  line  through  the  vertex  of  a  triangle 
and  parallel  to  the  base,  makes  equal  angles  with 
the  sides,  the  triangle  is  isosceles. 

Given :  Z.a  =  /.  x,  etc. 

7.  The  median  of  an  isosceles  triangle  is  per- 
pendicular to  the  base.     [Use  188  (3).] 

8.  The  bisectors  of   two  supplementary-adja- 
cent angles  are  perpendicular  to  each  other. 

Proof:   Z.40J5  +  Z50C  =  180°(?);  \£AOB  + 
\/.BOC  =  90°  (Ax.  3).  \£AOB  =  /.ROB;  etc. 

9.  The    bisectors    of    two    adjoining- interior 
angles  of  two  parallels  meet  at  right  angles. 

Proof:   ZMAC+  Z.MCA  =  90°  as  in  No.  8. 
Then  use  183. 


10.  If  the  bisector  of  an  exterior  angle  of  a  triangle 
is  parallel  to  the  base,  the  triangle  is  isosceles. 

11.  If  the  sum  of  two  angles  of  a  triangle  is  equal  to 
the  third,  the  triangle  is  a  right  triangle. 

[Use  110.]  B 


BOOK  I  65 

12.  If  the  median  of  a  triangle  is  equal  to  half  the  side  to  which  it  is 
drawn,  it  is  a  right  triangle. 

Given  :  MA  =  MB  =  MC. 
To  Prove  :  A  ABC  a  rt.  A. 
Proof  :  Z.4  =  Z  ACM  (?)  ;  ZB  =  ZBCM  (?). 
;.  by  adding,  etc.   (Use  Ax.  2  and  183.)  A 

13.  If  from  any  point  in  the  bisector  of  an  angle 
a  line  be  drawn  parallel  to  either  side  of  the  angle,  an 
isosceles  triangle  will  be  formed.    [182.] 

14.  If   the  bisector  of  the  vertex-angle  of  a  tri- 
angle is  perpendicular  to  the  base,  the  triangle  is 
isosceles. 

Proof:   Thert.  &are  =  [184  (4)].     Then  use  187 
(4);  etc. 

15.  Every  isosceles  right  triangle  can  be  divided 
by  one  line  into  two  isosceles  right  triangles. 


16.  The  diagonals  of  a  rhombus  divide  the  figure  into  four  equal 
right  triangles.     [141.] 

17.  If  a  line  be  drawn  perpendicular  to  the  bisec- 
tor of  an  angle  terminating  in  the  sides,  the  right 
triangles  formed  will  be  equaL 

18.  If  from  each  point  at  which  a  transversal  inter- 
sects two  parallels  a  perpendicular  to  the  other  paral- 
lel be  drawn,  two  equal  right  triangles  will  be  formed. 

19.  If  two    perpendiculars   be   drawn   to    the 

upper  base  of  a  parallelogram  from  the  extremi-       R    p T    C 

ties  of  the  lower  base,  two  equal  right  triangles 
will  be  formed. 

20.  The  perpendiculars  to  the  equal  sides  of  an 
isosceles  triangle  from  the  opposite  vertices  form 
two  pairs  of  equal  right  triangles. 


21.   If  two  intersecting  lines  have  their  extremities      — 
two  parallels  and  their  point  of  intersection  bisects 
le  of  them,  it  bisects  the  other  also. 
Given:   AO  =  E0\  etc.  c 


66 


PLANE   GEOMETRY 


22.  If  two  adjacent  sides  of  a  quadrilateral  are 
equal   and  the    diagonal   bisects    their    included 
angle,  the  other  two  sides  are  equal. 

23.  If  a  diagonal  of  a  quadrilateral  bisects  two 
of   its  angles,  the  quadrilateral  has  two  pairs  of 
equal  sides. 

24.  The  altitudes  of  an  isosceles  triangle  upon  the  legs  are  equal. 

25.  If  a  triangle    has  two  equal  altitudes,   it   is 
isosceles. 

26.  The  diagonals  of  a  rectangle  are  equal. 

27.  The  medians  drawn  from  the  ends  of  the  base  of 
an  isosceles  triangle  are  equal. 

28.  The   three  lines  joining  the  midpoints  of  the 
sides  of  a  triangle   divide  the  triangle  into  four 
equal  triangles.     [Use  142  and  132.] 

29.  The     bisector    of   the   vertex-angle  of    an 
isosceles  triangle  bisects  the  base  at  right  angles. 
[Use  185(1);  27;  50.] 

30.  The  bisectors  of  the  equal  angles  of  an  isos- 
celes triangle  (terminating  in  the  equal  sides)  are  equal.     [Use  185  (2).] 

31.  The  median  to  the  base  of  an  isosceles  triangle  bisects  the  vertex- 
angle.     [Use  185  (3).] 

32.  The  diagonals  of  an  isosceles  trapezoid  are  ^ 
equal.     [Use  145.] 

33.  The  diagonals  of  an  isosceles  trapezoid  divide 
the  figure  into  four  triangles,  of  which  one  pair  is 
isosceles  and  the  other  pair  is  equal. 


34.  What    is  the   complement   of  an  angle  containing    35°?    80°? 
75°  25'  ?  8°  18'  ? 

35.  What  is  the  supplement  of  50°?  100°?   148°?    113°  48'  ? 

36.  In  a  right  triangle  ABC,  if  Z  A  is  47°,  find  Z  B. 

37.  In  an  isosceles  triangle  Z  A  =  Z  B  =  80° ;  find  /.  C. 

38.  In  A  ABC,  if  Z  A  =  25°,  Z  B  =  88°,   find  Z  C.    Find  the  exterior 
angle  at  A . 


BOOK  I  67 

39.  In  A  ABC,  if  /.A  =  40°,  Z  B  =  70°  40' ;  find  Z  C  and  the  exterior 
angle  at  B. 

40.  The  vertex-angle  of  an  isosceles  triangle  is  44°.    Find  each  base 
angle. 

41.  How  many  degrees  are  there  in  the  sum  of  the  angles  of  a  penta- 
gon ?   of  a  decagon  ?  of  a  9-gon  ? 

42.  How  many  degrees  are  there  in  each  angle  of  an  equiangular  hexa- 
gon ?  of  an  equiangular  octagon  ? 

43.  How  many  degrees  are  there  in  each  exterior  angle  of  an  equian- 
gular pentagon  V    hexagon?    dodecagon?    16-gon? 

44.  If  one  acute  angle  of  a  right  triangle  is  double  the  other,  how 
many  degrees  are  there  in  each  ?    [Denote  the  less  Z  by  x.~\ 

45.  If  the  acute  angles  of   a  right  triangle  are  equal,  how   many 
degrees  are  there  in  each  ? 

46.  If  one  acute  angle  of  a  right  triangle  is  five  times  the  other,  how 
many  degrees  are  there  in  each  ? 

47.  If  a  base  angle  of  an  isosceles  triangle  is  60°,  find  the  vertex-angle. 
What  kind  of  triangle  is  this  ? 

48.  If  the  vertex-angle  of  an  isosceles  triangle  is  60°,  find  each  base 
angle.     What  kind  of  triangle  is  this  ? 

49.  If  the  vertex-angle  of  an  isosceles  triangle  equals  twice  the  sum 
of  the  two  base  angles,  how  many  degrees  are  there  in  each  angle? 

50.  If  the  vertex-angle  of  an  isosceles  triangle  equals  four  times  the 
sum  of  the  base  angles,  find  each  angle. 

51.  If  the  vertex-angle  of  an  isosceles  triangle  is  half  each  of  the  base 
angles,  find  each  angle. 

52.  If  one  angle  of  a  parallelogram  is  54°,  how  many  degrees  are 
there  in  each -of  the  remaining  angles? 

53.  If  a  transversal  cuts  two  parallels  making  one  pair  of  alternate- 
interior  angles  each  40°,  how  many  degrees  are  there  in  each  of  the  other 
six  angles  formed  ? 

54.  Find  the  three  angles  formed  by  the  bisectors  of  the  angles  of  a 
triangle  whose  angles  are  44°,  62°,  and  74°. 

55.  If  z  A  of  A  ABC  is  33°  and  the  exterior  angle  at  C  is  110°, 
find  Z  B. 


PLANE  GEOMETRY 


56.  If  two  angles  of  a  triangle  are  80°  and  55°,  how  many  degrees  are 
there  in  the  angle  formed  by  their  bisectors? 

57.  The  vertex-angle  of  an  isosceles  triangle  is  one  third  of  either  ex- 
terior angle  at  the  extremities  of  the  base.     Find  each  angle  of  the 
triangle. 

58.  If  two  angles  of  a  triangle  are  30°  and  40°, 
how  many  degrees  are  there  in  the  angle  formed 
by  the  bisector  of  the  third  angle  and  the  altitude 
from  the  same  vertex  ?     Solution :  Z  x  =  Z  A  BS  — 
Z  ABD  =  \Z  ABC  -  comp.  of  Z  A. 

59.  Theorem.    The  angle  between  the  altitude  of  a  triangle  and  the 
bisector  of  the  angle  at  the  same  vertex  equals  half  the  difference  of  the 
other  angles  of  the  triangle. 

Proof:    Z  x  =  Z  ABS  -  Z  ABD  =  \   (180°  - 
ZA-  Z  C)-  (90°  -  Z  A)  =  etc. 

60.  Theorem.    The  exterior  angle  at  the  base 
of  an  isosceles  triangle  equals  half  the  vertex- 
angle  plus  90°.    Proof :  Zx  =  Za  +  Zr  =  etc. 

61.  If  in  A  ABC,  Z  BAG  =  80°,  Z  ABC  =  30°, 
find  the  angle  formed  by  the  bisectors  of  the  exte- 
rior angles  at  A   and  B.     Solution  :  Z  x  =  180°  — 
ZBAD-ZABD;    Z  BAD  =  $  (180°- Z  A);   etc. 

62. 

terior  angles  of  a  triangle  equals 

the  interior  angles  at  the  same  vertices. 

63.  If  one  acute  angle  of  a  right  triangle  is  double 
the  other,  the  hypotenuse  is  double  the  shorter  leg. 

[Denote  the   less  Z  by  x.     Find  the  other.     Draw  the  median  from 
vertex  of  rt,  Z.     Prove  one  A  formed  equilateral.] 

64.  If  one  angle  of  a  triangle  is  double  another, 
the  line  from  the  third  vertex,  making  with  the 
longer  adjacent  side   an    angle  equal  to  the  less 

given  angle,  divides  the  triangle  into  two  isosceles  triangles. 


The  angle  formed  by  the  bisectors  of  two  ex-     •/ 
anles  of  a  trianle  euals  half  the  sum  of  ' 


65.   The  bisector  of  an  angle  bisects,  if  pro-     „ 
duced,  the  vertical  angle  also. 

Given :  ZDON=Z  BON.  c 


BOOK  I 


69 


66.  A  line  perpendicular  to  the  bisector  of   an  angle  at  the  vertex 
bisects  its  supplementary-adjacent  angle.  _ 

Given:  Z  a=  Z  b  and  OY  JL  to  ON. 
(Use  48.) 

67.  If  the  bisectors  of  two  adjacent  angles 
are  perpendicular,  the  angles  are  supplementary. 

Given :  Z  a  =  Z  £> ;  Z.x  —  Zz ;  Z NO Y  =  rt. Z. 

68.  The  bisectors  of  any  two  adjoining  angles  of  a 
parallelogram  meet  at  right  angles.      [Use  136 ;  183.] 

69.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  perpendiculars  to  the  equal  sides  be  drawn, 
they  will  make  equal  angles  with  the  base. 

[Use  114;  48.] 

70.  If  a  line  be  drawn  through  the  vertex  of  an 
angle  and  perpendicular  to  the  bisector  of  the  angle, 
it  will  make  equal  angles  with  the  sides. 

To  Prove :  Z  r  =  Z  s. 

71.  The  bisector  of  the  exterior  angle  at  the  ver- 
tex of  an  isosceles  triangle  is  parallel  to  the  base. 

Proof :   Z  DCB  =  2  Z  A   (?)  and  =  2  Z  DCR  (?). 
Etc. 

72.  The  line  through  the  vertex  of  an  isosceles  tri- 
angle, parallel  to  the  base,  bisects  the  exterior  angle. 

73.  Parallel  lines  are  everywhere  equally  distant. 
Given :  ||.  A  C  and  BD ;  A  B  and  CD  _fc  to  A  C. 

To  Prove :  AB  =  CD.     (Use  93;  130.) 

74.  If  two  lines  in  a  plane  are  everywhere  equally 
distant,  they  are  parallel.     [Use  93  ;  135.] 

75.  If  the  diagonals  of  a  parallelogram  are  equal, 
the  figure  is  a  rectangle. 

[Use  &  ABC  and  XXRC;  185  (3);  50.] 

76.  The  perpendiculars  upon  a  diagonal  of  a 
parallelogram    from    the    opposite    vertices    are 
equal.     [184  (1).]  A* 


t 


70  PLANE   GEOMETRY 

77.  The  perpendiculars  to  the  legs  of  an  isosceles  triangle  from  the 
midpoint  of  the  base  are  equal. 

78.  State  and  prove  the  converse  of  No.  77. 

79.  If  AB  =  LM  and  AL  =  BM,  ZB  =  ZL 
and  Z  BA 0  =  Z.  OML  and  BO=OL. 

80.  Any  line  terminated  in  a  pair  of  oppo- 
site   sides    of    a    parallelogram    and    passing 
through  the  midpoint  of  a  diagonal  is  bisected 
by  this  point.    To  Prove :  RO=OS. 

81.  The  midpoint  of  a  diagonal  of  a  paral- 
lelogram  is  a  center  of  symmetry. 

82.  If  the  base  angles  of  a  triangle  be  bisected 
and  through  the  intersection  of  the  bisectors  a  line 
be  drawn  parallel  to  the  base  and  terminating  in  the 
sides,  this  line  will  be  equal  to  the  sum  of  the  parts 

of  the  sides  it  meets,  between  it  and  the  base.  A  ~~*C 

83.  In  two  equal  triangles,  homologous  medians  are  equal.     Homolo- 
gous altitudes  are  equal.     Homologous  bisectors  are  equal. 

84.  If  two  parallel  lines  are  cut  by  a  transversal,  the  two  exterior 
angles  on  the  same  side  of  the  transversal  are  supplementary. 

85.  If  from  a  point  a  perpendicular  be  drawn  to  each  of  two  parallels 
they  will  be  in  the  same  line.     [Draw  a  third  II  through  the  point.] 


86.  One  side  of  a  triangle  is  less  than  the  sum  of  the  other  two  sides. 

87.  The  sum  of  the  sides  of  any  polygon  ABODE  is  greater  than  the 
sum  of  the  sides  of  triangle  A  CE. 

88.  If  X  is  a  point  in  side  AB  of  A  ABC,  A  B  -f  BO  A X  +  XC. 

89.  In  the  figure  of  No.  88  Z  AXO  Z  B. 

90.  If  lines  be  drawn  from  any  point  within  a  triangle  to  the  ends  of 
the  base,  they  will  include  an  angle  which  is  greater  than  the  vertex 
angle  of  the  triangle.     [Use  109  with  figure  of  75.] 

91.  Any  point  (except  the  vertex)  in  either  leg  of  an  isosceles  triangle 
is  unequally  distant  from  the  ends  of  the  base.  Q 

92.  If  two  sides  of  a  triangle  are  unequal  and 
the  median  to  the  third  side  be  drawn,  the  angles 
formed  with  the  base  will  be  unequal.     [Use  87.] 

93.  State  and  prove  the  converse  of  No.  92. 


BOOK    I 


71 


94.  If  the  side  LM,  of  equilateral  triangle  LMN, 
be  produced  to  P,  and  PN  be  drawn,  Z  PNL  >  Z  L  > 
Z  P.     Also  PL>PN>  LN. 

95.  If  from  any  point  within  a  triangle  lines  be 
drawn  to  the  three  vertices: 

(1)  Their  sum  will  be  less  than  the  sum  of  the  sides 
of  the  triangle.     [Use  75  three  times.] 

(2)  Their  sum  will  be  greater  than  half  the 
sum  of  the  sides  of  the  triangle.     [Use  Ax.  12 
three  times.] 

96.  The  sum  of  the  diagonals  of  any  quadri- 
lateral is  less  than  the  sum  of  the  four  sides; 
but  greater  than  half  that  sum. 

97.  The  line  drawn   from  any  point  in  the  base  of   an   isosceles 
triangle  to  the  opposite  vertex  is  less  than  either  leg. 


98.  The  bisectors  of  a  pair  of  corresponding  angles  are  parallel. 
[Use  98  ;  189,  etc.] 

99.  If  two  lines  are  cut  by  a  transversal  and  the  exterior  angles  on 
the  same  side  of  the  transversal  are  supplementary,  the  lines  are  parallel. 

100.  The  bisectors  of  a  pair  of   vertical   angles  are   in    the  same 
straight  line. 

101.  If  one  angle  of  a  parallelogram  is  a  right  angle  the  figure  is  a 
rectangle. 

102.  The  bisectors  of  the  angles  of  a  trapezoid  form  a  quadrilateral 
two  of  whose  angles  are  right  angles. 

103.  The  bisectors    of    the  four  interior 
angles   formed   by  a  transversal  cutting  two 
parallels  form  a  rectangle. 

[Prove  each  Z  of  LMPQ  a  rt.  Z.] 

104.  The  bisectors  of  the  angles  of  a  par- 
allelogram form  a  rectangle. 

105.  The  bisectors  of  the  angles  of  a  rectangle 
form  a  square.     [In  order  to  prove  EFGH  equilat- 
eral, the  &  A  HE  and  CDF  are  proved  equal  and 
isosceles;  similarly  &BGC  and  AED.~\ 


72 


PLANE   GEOMETRY 


106.  The    lines  joining  a  pair  of   opposite 
vertices  of  a  parallelogram  to  the  midpoints  of 
the  opposite  sides  are  equal  and  parallel.    [Prove 

,  BCEF'a.a.'] 

107.  If  the  four  midpoints  of  the  four  halves 
of  the  diagonals  of  a  parallelogram  be  joined 
in  order,  another  parallelogram  will  be  formed. 

108.  If  the  points  at  which  the  bisectors 
of   the   equal  angles  of   an  isosceles  triangle 
meet  the  opposite  sides,  be  joined  by  a  line, 
it  will  be  parallel  to  the  base. 

109.  If  two  angles  of  a   quadrilateral  are   supple- 
mentary, the  other  two  are  supplementary.    [Use  165.] 

110.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  parallels  to  the  equal  sides  be  drawn,  the  sum 
of  the  sides  of  the  parallelogram  formed  will  be  equal 
to  the  sum  of  the  legs  of  the  triangle. 

To  Prove:  XY+  YC  +  CZ  +  XZ  =  AC  +  BC. 

111.  If  one  of  the  legs  of  an  isosceles  triangle  be 
produced  through  the  vertex  its  own  length,  and  the 
extremity  be  joined  to  the  nearer  end  of  the  base, 
this  line  will  be  perpendicular  to  the  base.    [Use  183.] 

112.  If  the  middle  point  of  one  side  of  a  triangle 
is  equally  distant  from  the  three  vertices,  the  triangle 
is  a  right  triangle. 

[Proof  and  figure  same  as  for  No.  111.] 

113.  If  through  the  vertex  of  the  right 
angle  of   a  right  triangle  a  line  be  drawn 
parallel  to  the  hypotenuse,  the  legs  of  the 
right  triangle  will  bisect  the  angles  formed 
by  this  parallel   and  the  median  drawn  to 
the  hypotenuse.    [Use  148 ;  97 ;  etc.]  A 

114.  Any  two   vertices    of    a    triangle    are 
equally  distant  from  the  median  from  the  third 
vertex. 

115.  If  from  any  point  within  an  angle  per- 
pendiculars  to  the   sides  be  drawn,  they  will 
include  an  angle  which  is  the  supplement  of 
the  given  angle. 


A 


M 


BOOK  I 


73 


116.  The  lines  joining   (in  order)   the  midpoints  of  the  sides  of  a 
quadrilateral  form   a  parallelogram   the   sum   of 

whose  sides  is  equal  to  the  sum  of  the  diagonals  of 
the  quadrilateral.      [Use  142.] 

117.  The    lines   joining   (in   order)    the    mid- 
points of  the  sides  of  a  rectangle  form  a  rhombus. 
[Draw  the  diagonals.]  A 

118.  If  a  perpendicular  be  erected  at  any  point  in 
the  base  of  an  isosceles  triangle,  meeting  one  leg,  and 
the  other  leg  produced,  another  isosceles  triangle  will 
be  formed. 

[Z    o  and   Z.  S   are     complements  of  =  A  A   and 
B  (?).    Etc.] 

119.  The  difference  between   two  sides    of   a 
triangle  is  less  than  the  third  side. 

120.  The  bisectors  of  two  exterior  angles  of  a 
triangle  and  of  the  interior  angle  at  the  third  ver- 
tex meet  in  a  point. 

121.  The  bisectors  of  the  exterior  angles  of  a 
rectangle  form  a  square. 

122.  If  lines  be  drawn  from  a  pair  of  oppo- 
site vertices  of  a  parallelogram  to  the   mid- 
points of  a  pair  of  opposite  sides,  they  will 
trisect  the  diagonal  joining  the  other  two  ver- 
tices.    [Prove  AECFa,  £7  and  use  143  in 
DYCandABX.] 

123.  If  two  medians  of  a  triangle  are  equal, 
the  triangle  is  isosceles. 

[Use  151.  AO  =  OB  (Ax.  3).     Hence  prove  A 
AEO  and  DEO  equal.] 

124.  How  many  sides  has  the  polygon  the  sum 
of  whose  interior  angles  exceeds  the  sum  of  its 
exterior  angles  by  900  °  ? 

125.  If  the  vertex-angle  of  an  isosceles  triangle 
is  twice  the  sum  of  the  base  angles,  any  line  per- 
pendicular to  the  base  forms  with  the  sides  of  the 
given    triangle    (one    side  to    be    produced)    an 
equilateral  triangle.     [Use  121.] 


74 


PLANE   GEOMETRY 


126.  The  lines  bisecting  two  interior  angles  that  a  transversal  makes 
with  one  of  two  parallels  cut  off  equal  segments  on  the  other  parallel  from 
the  point  at  which  the  transversal  meets  it.    [The  &  formed  are  isosceles.] 

127.  The  bisector  of  the  right  angle   of  a 
right  triangle  is  also  the  bisector  of  the  angle 
formed  by  the  median  and  the  altitude  drawn 
from  the  same  vertex. 

To  Prove :  Z  MCS  =  LCS.  A 

Proof:   /.ACS=  ^BCS  (?);  ZACM  =  Z  BCL  (?).   Now  use  Ax.  2. 

128.  If  through  the  point  of  intersection  of  the  diagonals  of  a  par' 
allelogram,  two  lines  be  drawn   intersecting   a  pair  of  opposite  sides 
(produced  if  necessary),  the  intercepts  on  these  sides  will  be  equal. 

129.  If  ABC  is  a  triangle,  BS  is  the  bisector 
of  /.  ABC,  and  A M  is  parallel  to  BS  meeting  EC 
produced,  at  M,  the  triangle  ABM  is  isosceles. 

130.  If  AB C  is  a  triangle    and   BS  is   the 
bisector  of  exterior  Z  ABR  and  AM  is  II  to  BS 
meeting  BC  at  M,  A  ABM  is  isosceles. 

131.  If  A,  B,  C,  and   D  are  points  on  a 
straight  line  and  AB  =  BC,  the  sum  of  the 
perpendiculars  from  A  and  C  to  any  other 
line  through  D  is  double  the  perpendicular 
to  that  line  from  B.    [Use  147  ;  144.] 

132.  If  on  diagonal  BD,  of  square  A  BCD,  BE  be 
taken  equal  to  a  side  of  the  square,  and  EP  be  drawn 
perpendicular  to  BD  meeting  AD  at  P,  AP  =  PE  = 
ED.     [Draw  BP.~] 

133.  If  in  A  ABC,  Z  A  is  bisected  by  line  meet- 
ing BC  at  M,  AB>BM  and  AC>  CM.      [Use  109; 
123.] 

134.  It  is  impossible  to  draw  two  straight  lines 
from  the  ends  of  the  base  of  a  triangle  terminating 
in  the  opposite  side,  so  that  they  shall  bisect  each 
other.     [Use  138.] 

135.  If   ABC  is   an  equilateral  triangle   and 
D,  E,  F  are  points  on  the  sides,  such  that  AD  = 
BE  =  CF,  triangle  DEF  is  also  equilateral.  t 

[Prove  the  three  small  A  = .]  A 


BOOK  I 


75 


136.  If  ABCD  is  a  square  and  E,  F,  G,  H  are 
points  on  the   sides,  such  that   AE  =  BF  =  CG  = 
DH,  EFGH  is  a  square. 

[First,  prove  EFGH  equilateral ;  then  one  Z  a  rt.  Z .] 

137.  If  ABC  is  an  equilateral  triangle  and  each 
side  is  produced  (in  order)  the  same  distance,  so  that 
AD  =  BE=CF,  the  triangle  DEF  is  equilateral. 

138.  If  ABCD  is  a  square  and  the  sides  be  produced  (in  order)  the 
same  distance,  so  that  AE  =  BF  =  CG  =  DH,  the  figure  EFGH  will  be 
a  square. 

139.  The  two  lines  joining  the  midpoints  of  the  opposite  sides  of  a 
quadrilateral  bisect  each  other.     [Join  the  4  midpoints  (in  order),  etc.] 

140.  If  twp  adjacent  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  other  angles  are  perpendicular  to  each  other. 

141.  If  two  opposite  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  otjher  angles  are  parallel. 

142.  Two  isosceles  triangles  are  equal,  if: 

(1)  The  base  and  one  of  the  adjoining  angles  in  the  one  are  equal 
respectively  to  the  base  and  one  of  the  adjoining  angles  in  the  other. 

(2)  A  leg  and  one  of  the  base  angles  in  the  one  are  equal  respectively 
to  a  leg  and  one  of  the  base  angles  in  the  other. 

(3)  The  base  and  vertex-angle  in  one  are  equal  to  the  same  in  the 
other. 

(4)  A  leg  and  vertex-angle  in  one  are  equal  to 
the  same  in  the  other. 

(5)  A  leg  and  the  base  in  one  are  equal  to  the 
same  in  the  other. 

143.  If  upon  the  three  sides  of  any  triangle 
equilateral   triangles  be   constructed  (externally) 
and  a  line  be  drawn  from  each  vertex  of  the  given 
triangle  to  the  farthest  vertex  of  the  opposite  equi- 
lateral triangle,  these  three  lines  will  be  equal. 

Proof:  ZEAC  =  Z  BAF  (?).     Add  to  each  R 
of  these,  Z  CAB.    :.  Z  EAB  =  Z  CA  F  (?). 
Then  prore  &  EAB  and  CAF  equal. 
Similarly,  A  CA  D  =  A  CEB.     Etc. 

144.  If   two    medians  be   drawn   from   two 
vertices  of  a  triangle  and  produced  their  own 
length  beyond  the  opposite  sides  and  these  ex- 
tremities be  joined  to  the  third  vertex,  these  two 

lines  will  be  equal,  and  in  the  same  straight  line.    [Draw  MP  and  use  142.] 


76  PLANE   GEOMETRY 

145.  The  median  to  one  side  of  a  triangle  is  less  than  half  the  sum  ot 
the  other  two  sides. 

Proof:  (Fig.  of  No.  144.)  Produce  median  EM  its  own  length  to  #, 
draw  RA.     Prove  RA  -  CB.     Prove  BR  <  AB  +  EC,  etc. 

146.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  sum  of  the 

sides  of  the  triangle.  r->      _  C 

/\-  ^^\ 

147.  If  the  diagonals  of  a  trapezoid  /* 

are  equal,  it  is   isosceles. 

[Draw  DR  and  CS  _L  to  AB  ;  and 
prove  rt.  A  ACS  and  BDR  equal,  to 
get  Z.  x  = 


148.  If  a  perpendicular  be  drawn  from  each  vertex  of  a  parallelogram 
to  any  line  outside  the  parallelogram,  the  sum  of  those  from  one  pair  of 
opposite  vertices  will  equal  the  sum  of  those  from  the  other  pair. 

[Draw  the  diagonals  ;  use  144.] 

149.  The  sum  of  the  perpendiculars  to  the  legs 
of  an  isosceles  triangle   from   any  point    in   the 
base  equals  the  altitude   upon   one   of  the  legs. 
(That  is,  the  sum  of  the  perpendiculars  from  any 
point  in  the  base  of  an  isosceles  triangle  to  the  equal 
sides  is  constant  for  every  point  of  the  base.) 

[Prove  PE  =  CF  by  184  (1).]  A         P~         ~c 

150.  The  sum  of  the  three  perpendiculars  drawn  from  any  point 
within  an  equilateral  triangle,  to  the  three  sides,  is  constant  for  all 
positions  of  the  point. 

[Draw  a  line  through  this  point  II  to  one  side;  draw  the  altitude  of  the 
A  J.  to  this  line  and  side  ;  prove  the  sum  of  the 
three  Js  =  this  altitude  and  hence,  =  a  constant.] 

151.  The  line  joining  the  midpoints  of  one 
pair  of  opposite  sides  of  a  quadrilateral  and  the 
line  joining  the    midpoints   of    the   diagonals 
bisect  each   other. 


To  Prove  :  LM  and  RS  bisect  each  other. 

152.  If  one  leg  of  a  trapezoid  is  perpendicular  to  the  bases,  the  mid- 
point of  the  other  leg  is  equally  distant  from  the  ends  of  the  first  leg. 
[Draw  the  median.] 

153.  The  median  of  a  trapezoid  bisects  both  the  diagonals. 

154.  The  line  joining  the  midpoints  of  the  diagonals  of  a  trapezoid  is 
a  part  of  the  median,  is  parallel  to  the  bases,  and  is  equal  to  half  their 
difference. 


BOOK  I  77 

155.  If,  in  isosceles  triangle  X  YZ,  AD  be  drawn  from  A,  the  midpoint 
of  YZ,  perpendicular  to  the  base  XZ,  DZ  =  $  XZ.    [Draw  alt.  from  7.] 

156.  If  ABC  is  an  equilateral  triangle,  the  bisectors  of  angles  B  and  C 
meet  at  Z>,  DE  be  drawn  parallel  to  AB  meeting  AC  at  E,  and  DF, 
parallel  to  EC  meeting  A  C  at  F,  then  AE  =  ED  =  EF=  DF=  CF. 

157.  If  A  is  any  point  in  RS  of  triangle  RST,  and  £  is  the  midpoint 
of  RA,  C  the  midpoint  of  AS,  D  the  midpoint  of  ST,  and  E  the  mid- 
point of  TR,  then  BCDE  is  a  parallelogram. 

158.  In  a  trapezoid  one  of  whose  bases  is 
double  the  other,  the  diagonals  intersect  at  a 
point  two  thirds  of  the  distance  from  each  end 
of  the  longer  base  to  the  opposite  vertex. 

Proof:  Take  M,  the  midpoint  of  AO,  etc. 

159.  If  lines  be  drawn  from  any  vertex  of  a  parallelogram  to  the  mid- 
points of  the  two  opposite  sides,  they  will  divide  the  diagonal  which  they 
intersect,  into  three  equal  parts. 

Proof :   Draw  the  other  diagonal  and  use  151. 

160.  If  the  interior  and  exterior  angles  at  two  vertices  of  a  triangle 
be  bisected,  a  quadrilateral  will  be  formed,  two  of  whose  angles  are  right 
angles  and  the  other  two  are  supplementary. 

161.  The  angle  between  the  bisectors  of  two  angles  of  a  triangle 
equals  half  the  third  angle  plus  a  right  angle. 

162.  If,  in  triangle  ABC,  the  bisectors  of  the  interior  angle  at  B  and 
of  the  exterior  angle  at  C,  meet  at  Z>,  the  angle  BA  C  equals  twice  the 
angle  BDC. 

163.  The  four  bisectors  of  the  angles  of  a  quadrilateral  form  a  second 
quadrilateral  whose  opposite  angles  are  supplementary. 

Proof:  Extend  a  pair  of  opposite  sides  of  the  given  quadrilateral 
to  meet  at  X.  Bisect  the  base  angles  of  the  new  A  formed,  meeting  at  0. 
Then  show  that  Z.  O  equals  one  of  the  A  between  the  given  bisectors, 
and  Z  O  is  supplementary  to  the  angle  opposite.  . 

164.  The  sum  of  the  angles  at  the  vertices  of  a  A 
five-pointed  star  (pentagram)  is  equal  to  two  right                /  \ 
angles. 

Proof :  Draw  interior  pentagon.     Find  number  of 
degrees  in  each  of  its  angles.     Hence  find  Z  A,  etc. 

165.  The  lines  joining  the  midpoints  of  the  op- 
posite sides  of  an  isosceles  trapezoid  are  perpendicular  t6  each  other. 


78  PLANE   GEOMETRY 

166.  If  the  opposite  sides  of  a  hexagon  are  equal  and  parallel,  the 
three  diagonals  drawn  between  opposite  vertices  meet  in  a  point. 

167.  In  triangle  ABC,  AD  is  perpendicular  to  BC,  meeting  it  at  Z>; 
E  is  the  midpoint  of  AB,  and  F  of  A  C;  the  angle  EDF  is  equal  to  the 
angle  EAF.     [Use  148;  55.] 

168.  If  the  diagonals  of  a  quadrilateral  are  equal,  and  also  one  pair 
of  opposite  sides,  two  of  the  four  triangles  into  which  the  quadrilateral 
is  divided  by  the  diagonals  are  isosceles. 

169.  If  angle  A  of  triangle  ABC  equals  three  times  angle  B,  there 
can  be  drawn  a  line  AD  meeting  BC  in  D,  such  that  the  triangles  ABD 
and  A  CD  are  isosceles. 

170.  If  E  is  the  midpoint  of  side  BC  of  parallelogram  A  BCD.  AE 
and  BD  meet  at  a  point  two  thirds  the  distance  from  A  to  E  and  from 
Dio  B. 

171.  If  in  triangle  ABC,  in  which  AB  is  not  equal  to  AC,  A  C'  be 
taken  on  AB  (produced  if  necessary)  equal  to  A  C^nd  AB'  be  taken 
on  AC  (produced  if  necessary)  equal  to  AB,  and  B'C'  be  drawn  meeting 
BC&tD,  then  AD  will  bisect  angle  BA  C. 

Proof:  A  ABC  =  A  A  B'C'  (?)  (52).     /.  their  homologous  parts  are 
equal.     Thus  prove  ABC'D  =  A  B'CD  (54).     Etc. 

172.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  it  also  bisects 
the  opposite  angle. 

173.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  the  figure  is 
equilateral. 

174.  Any  line  drawn  through  the  point  of  intersection  of  the  diago- 
nals of  a  parallelogram  divides  the  figure  into  two  equal  trapezoids. 
[See  159.] 

175.  If  AR  bisects  angle  A  of  triangle  ABC  and  A  T  bisects  the  ex- 
terior angle  at  A,  any  line  parallel  to  AB,  having  its  extremities  in  AR 
and  A  T,  is  bisected  by  A  C. 

176.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the  figure  is 
a  parallelogram.     [See  165.] 


BOOK   II 

THE  CIRCLE 

191.  A  curved  line  is  a  line  no  part  of  which  is  straight. 

192.  A  circumference  is  a  curved  line  every  point  of  which 
is  equally  distant  from  a  point  within,  called  the  center. 

193.  A  circle  is  a  portion  of  a  plane  bounded  by  a  cir- 
cumference.    [O.] 

194.  A  radius  is  a  straight  line  drawn  from  the  center  to 
the  circumference. 

A  diameter  is  a  straight  line  containing  the  center,  and 
whose  extremities  are  in  the  circumference. 


CIRCUMFERENCE                        SECANT                  CENTRAL    ANGLE  SEMI- 
CIRCLE                                    CHORD                  INSCRIBED    ANGLE  CIRCUMFERENCES 
RADIUS                                TANGENT                              ARC                                  SEMICIRCLES 
DIAMETER  POINT   OF    CONTACT 

A  secant  is  a  straight  line  cutting  the  circumference  in  two 
points. 

A  chord  is  a  straight  line  whose  extremities  are  in  the  cir- 
cumference. 

A  tangent  is  a  straight  line  which  touches  the  circumference 
at  only  one  point,  and  does  not  cut  it,  however  far  it  may  be 
extended.  The  point  at  which  the  line  touches  the  circum- 
ference is  called  the  point  of  contact  or  the  point  of  tangency. 

79 


80  PLANE   GEOMETRY 

195.  A  central  angle  is  an  angle  formed  by  two  radii. 

An  inscribed  angle  is  an  angle  whose  vertex  is  on  the  cir- 
cumference and  whose  sides  are  chords. 

196.  An  arc  is  any  part  of  a  circumference. 

A  semicircumference  is  an  arc  equal  to   half  a   circum- 
ference. 

A  quadrant  is  an  arc  equal  to  one  fourth  of  a  circumference. 
Equal  circles  are  circles  having  equal  radii. 
Concentric  circles  are  circles  having  the  same  center. 


CIRCLES    INTERNALLY        CIRCLES    EXTERNALLY 
TANGENT  TANGENT 


197.  A  sector  is  the  part  of  a  circle  bounded  by  two  radii 
and  their  included  arc. 

A  segment  of  a  circle  is  the  part  of  a  circle  bounded  by  an 
arc  and  its  chord. 

A  semicircle  is  a  segment  bounded  by  a  semicircumference 
and  its  diameter. 

198.  Two  circles  are  tangent  to  each  other  if  they  are  tan- 
gent to  the  same  line  at  the  same  point.      Circles  may  be 
tangent   to   each   other  internally,  if  the  one  is  within  the 
other,  or  externally,  if  each  is  without  the  other. 

199.  POSTULATE.     A   circumference  can  be  described  about  any 
given  point  as  center  and  with  any  given  line  as  radius. 

Explanatory.  A  circle  is  named  either  by  its  center  or  by 
three  points  on  its  circumference,  as  "the  O  O,"  or  "the  O 
ABC." 

The  verb  to  subtend  is  used  in  the  sense  of  "to  cut  off." 


BOOK  II  81 

A  chord  subtends  an  arc.  Hence  an  arc  is  subtended  by 
a  chord. 

An  angle  is  said  to  intercept  the  arc  between  its  sides. 
Hence  an  arc  is  intercepted  by  an  angle. 

The  hypothesis  is  contained  in  what  constitutes  the  sub- 
ject of  the  principal  verb  of  the  theorem.  (See  59.) 

PRELIMINARY  THEOREMS 

200.  THEOREM.   All  radii  of  the  same  circle  are  equal.    (See  192.) 

201.  THEOREM.   All  radii  of  equal  circles  are  equal.     (See  196.) 

202.  THEOREM.   The  diameter  of  a  circle  equals  twice  the  radius. 

203.  THEOREM.   All  diameters  of  the  same  or  equal  circles  are 
equal.  (Ax.  3.) 

204.  THEOREM.   The  diameter  of  a  circle  bisects  the  circle  and  the 
circumference. 

Given  :    Any  O  and  a  diameter. 

To  Prove  :  The  segments  formed  are  equal,  that  is,  the 
diameter  bisects  the  circle  and  the  circumference. 

Proof  :  Suppose  one  segment  folded  over  upon  the  other 
segment,  using  the  diameter  as  an  axis.  If  the  arcs  do  not 
coincide,  there  are  points  of  the  circumference  unequally 
distant  from  the  center.  But  this  is  impossible  (?)  (192). 

.*.  the  segments  coincide  and  are  equal  (?)   (28).      Q.E.D. 

205.  THEOREM.   With  a  given  point  as  center  and  a  given  line  as 
radius,  it  is  possible  to  describe  only  one  circumference.     (See  192.) 

That  is,  a  circumference  is  determined  if  its  center  and  radius  are 
fixed. 

NOTE.  The  word  "  circle  "  is  frequently  used  in  the  sense  of  "  cir- 
cumference." Thus  one  may  properly  speak  of  drawing  a  circle.  The 
established  definitions  could  not  admit  of  such  an  interpretation  save  as 
custom  makes  it  permissible. 


Ex.  Draw  two  intersecting  circles  and  their  common  chord.  Draw 
two  circles  which  have  no  common  chord.  Draw  figures  to  illustrate  all 
the  nouns  defined  on  the  two  preceding  pages. 


82  PLANE  GEOMETRY 


THEOREMS  AND  DEMONSTRATIONS 

206.   THEOREM.   In  the  same  circle   (or  in  equal  circles)  equal 
central  angles  intercept  equal  arcs. 


Given :    O  O  =  O  C  ;  Z  O  =  Z  G. 

To  Prove  :    Arc  AB  =  arc  LM. 

Proof :  Superpose  O  O  upon  the  equal  O  O,  making  Z  O 
coincide  with  its  equal,  Z  C.  Point  A  will  fall  on  L,  and 
point  B  on  M  (?)  (201). 

Arc  AB  will  coincide  with  arc  LM  (?)  (192). 

.-.  AB  arc  =  arc  LM  (?)  (28).  Q.E.D. 

207.  THEOREM.  In  the  same  circle  (or  in  equal  circles)  equal 
arcs  are  intercepted  by  equal  central  angles.  [Converse.] 

Given :    O  O  =  O  C ;  arc  AB  =  arc  LM. 

To  Prove :   Z  o  =  Z  C. 

Proof :  Superpose  O  O  upon  the  equal  O  C,  making  the 
centers  coincide  and  point  A  fall  on  point  L.  Then  arc  AB 
will  coincide  with  arc  LM  and  point  B  will  fall  on  point  M. 
(Because  the  arcs  are  =.) 

Hence  OA  will  coincide  with  Ci,  and  OB  with  CM  (?)  (39). 

.-.Zo  =  ZC(?)    (28).  Q.E.D. 

Ex.  1.   Can  arcs  of  unequal  circles  be  made  to  coincide?     Explain. 
Ex.  2.   If  two  sectors  are  equal,  name  the  several  parts  that  must  be 
equal. 


BOOK  II  83 

208.   THEOREM.    In  the  same  circle  (or  in  equal  circles) : 

I.   If  two  central  angles  are  unequal,  the  greater  angle  intercepts  the 
greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  intercepted  by  the 
greater  central  angle.   [Converse.] 


I.    Given:    OO  =  OC;    Z  LCM  >  Z  o. 
To  Prove :    Arc  LM  >  arc  AB. 

Proof :    Superpose  O  O  upon  O  O,  making  sector  AOB  fall 
in  position  of  sector  XCM,  OB  coinciding  with  CM. 
CX  is  within  the  angle  LCM  (Z  LCM  >  Z  o). 
Arc  AB  will  fall  upon  LM,  in  the  position  XM  (192). 
/.  arc  LM  >  arc  XM  (Ax.  5).     That  is,  arc  LM  >  arc  AB. 

Q.E.D. 

II.    Given:   (?).     To  Prove:   Z  LCM  >  Z  o. 
Proof :    The  pupil  may  employ  either  superposition,  as  in  I, 
or  the  method  of  exclusion,  as  in  87. 

NOTE.  Unless  otherwise  specified,  the  arc  of  a  chord  always  refers  to 
the  lesser  of  the  two  arcs.  If  two  arcs  (in  the  same  or  equal  circles)  are 
concerned,  it  is  understood  either  that  each  is  less  than  a  semicircumfer- 
ence,  or  each  is  greater. 

Ex.  1.  Two  sectors  are  equal  if  the  radii  and  central  angle  of  one 
are  equal  respectively  to  the  radii  and  central  angle  of  the  other. 

Ex.  2.  If  in  the  figure  of  206,  arcs  AB  and  LM  were  removed,  how 
would  the  remaining  arcs  compare  ? 

Ex.  3.  If  in  the  figure  of  208,  arcs  AB  and  LM  were  removed,  how 
would  the  remaining  arcs  compare? 


84  PLANE   GEOMETRY 

209.   THEOREM.    In  the  same  circle   (or  in  equal  circles)  equal 
chords  subtend  equal  arcs. 


C 

Given  :    O  o  =  O  C ;  chord  AB  =  chord  LM. 
To  Prove  :    Arc  AB  =  arc  LM. 

Proof  :    Draw  the  several  radii  to  the  ends  of  the  chords. 
In  A  OAB  and  CLM,  OA  =  CL,  OB  =  CM  (?)   (201). 
Chord  AB  =  chord  LM  (Hyp,).  .'.A  OAB  =  A  CLM  (?). 
Hence  Z  O  =  Z  (7  (?). 
.'.  arc  AB  =  arc  LM  (?)   (206).  Q.E.D. 

210.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  equal  arcs 
are  subtended  by  equal  chords. 

Given  :    O  o  =  O  C  ;  arc  AB  =  arc  LM. 
To  Prove:    Chord  AB  =  chord  LM. 

Proof:    Draw  the  several  radii  to  the  ends  of  the  chords. 
In    A  OAB    and    CLM,   OA  =  CL,  OB  =  CM  (?)  (201). 
ZO=ZC(?)   (207).      /.A  AOB  =A  CLM  (?). 
.'.  chord  AB  =  chord  LM  (?).  Q.E.D. 

211.  THEOREM.  In  the  same  circle  (or  in  equal  circles)  : 

I.  If  two  chords  are  unequal,  the  greater  chord  subtends  the 
greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  subtended  by  the 
greater  chord. 

I.    Given  :    O  O  =  O  C  ;  chord  AB  >  chord  RS. 

To  Prove :    Arc  AB  >  arc  RS. 


BOOK  II 


85 


X—  R 

Proof:    Draw  the  several  radii  to  the  ends  of  the  chords. 
In  &AOB  and  RCS,  AO  =  RC,  BO  =  SC  (?)  (201). 

Chord  AB  >  chord  RS  (Hyp.).     /.Z  O  >  Z  C  (?)  (87). 

/.  arc  AB  >  arc  RS  (?)   (208,  I).  Q.E.D. 

II.    Given:  O  O  =  O  C;  arc  AB  >  arc  R8. 

To  Prove  :    Chord  AB  >  chord  RS. 

Proof :    Draw  the  several  radii.     In  A  AOB  and  RCS,  AO  = 
RC,  BO  =  SC  (?)  (201). 

But  Z  O  >  Z  c  (?)  (208,  II). 

.*.  chord  AB  >  chord  RS  (?)  (86).  Q.E.D. 

212.   THEOREM.  The  diameter  perpendicular  to  a  chord  bisects  the 
chord  and  both  the  subtended  arcs. 

Given  :  Diameter  DR  J_  to  chord 
AB  in  O  O. 

To  Prove:  I.  AM=MB;  II.  AR 

=  RB  and  AD  =  DB. 

Proof:    Draw  radii  to  the   ends 
of  the  chord. 

I.  In  rt.  A  OAM  and  OBM,  OA  = 
OB  (?),  OM=OM  (?). 

.'.  A  OAM=  A  OBJf  (?). 

Hence,  ^Jf=  J/B  (?).  Q.E.D. 

II.  Z ^OJf  =  Z  BOJf  (27).     .:AR  =  RB  (?)  (206). 

Also  Z^OD  =  Z  BOD  (?)(49).     .'.AD  =  DB  (?)(206).  Q.E.D. 


86 


PLANE   GEOMETRY 


213.  THEOREM.  The  line  from  the  center  of  a  circle  perpendicular 
to  a  chord  bisects  the  chord  and  its  arc.     Proof :  The  same  as  212. 

214.  THEOREM.   The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle.     [0  is  equidistant  from  A  and  B  (?)  (200). 
.'.  it  is  in  the  JL  bisector  of  AB  (?)  (69).] 

215.  THEOREM.  The  line  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  circle. 

Given:  Radius  OA  of 
O  O,  and  ET  _L  to  OA  at  A. 

To  Prove:  RT  tangent 
to  the  circle. 

Proof:  Take  any  point 
P  in  ET  (except  A)  and 
draw  OP. 

OP  >  OA  (?)  (77). 
Hence  P    lies    without  R 
the  O.     (Because  OP  >  radius.) 

That  is,  every  point  (except  A)  of  line  ET  is  without 
the  O. 

Therefore,  ET  is  a  tangent  (Def.  194).  Q.E.D. 

216.  THEOREM.   If  a  line  is  tangent  to  a  circle,  the  radius  drawn 
to  the  point  of  contact  is  perpendicular  to  the  tangent. 

Given  :    RT  tangent  to  O  O  at  A  ;  radius  OA. 
To  Prove  :    OA  J_  to  ET. 

Proof:  Every  point  (except  A)  in  RT  is  without  the  O 
(Def.  194). 

Therefore  a  line  from  O  to  any  point  of  RT  (except  .4) 
is  >  OA.  (Because  it  is  >  a  radius.)  That  is,  OA  is  t]ae 
shortest  line  from  o  to  RT.  .-.  OA  is  _L  to  RT  (?)  (77).  Q.E.D. 

217.  COR.  The  perpendicular  to  a  tangent  at  the  point  of  contact 
passes  through  the  center  of  the  circle.    (See  43.) 


BOOK  II 


87 


218.  THEOREM.  If  two  circles  are  tangent  to  each  other,  the  line 
joining  their  centers  passes  through  their  point  of  contact. 

Given :  ©  O  and  c 
tangent  to  a  line  at 
A,  and  line  OC. 

To  Prove :  OC 
passes  through  A. 

Proof:  Draw  radii 
OA  and  CA.  OA  is 

JL    to    the    tangent 

and  CA  is  _L  to  the  tangent   (?)   (216). 

.'.  OAC  is  a  st.  line  (?)  (43).     /.  OAC  and  OC.  coincide  and 
OC  passes  through  A  (39).  Q.E.D. 

Let  the  pupil  apply  this  proof  if  the  circles  are  tangent  internally. 

219.  THEOREM.  Two  tangents  drawn  to  a  circle  from  an  external 
point  are  equal. 


NOTE.     In  this  theorem  the  word  "  tangent "  signifies  the  distance  be- 
tween the  external  point  and  the  point  of  contact. 

Given:    OO;  tangents  P^L,  PB. 

To  Prove :    Distance  PA  =  distance  PB. 

Proof  :    Draw  radii  to  the  points  of  contact,  and  join  OP. 
A  OAP  and  OBP  are  rt.  A    (?)  (216). 

In    rt,   A  O^P    and    OBP,    OP  =  OP    (?)  ;    OA  =  OB    (•?). 
.'.  A  O^IP  =  A  OBP  (?).      .'•  PA  =  PB    (?).  Q.E.D. 


88 


PLANE   GEOMETRY 


220.    THEOREM.   If  from  an  external  point  tangents  be  drawn  to  a 
circle,  and  radii  be  drawn  to  the  points  of  contact,  the  line  joining  the 
center  and  the  external  point  will  bisect  : 
I.  The  angle  formed  by  the  tangents. 
II.  The  angle  formed  by  the  radii. 

III.  The  chord  joining  the  points  of  contact. 

IV.  The  arc  intercepted  by  the  tangents. 

Proof:  AoAPandOBP 
are  rt.  A  (?). 
They  are  =  .  (Explain.) 


II.  Z.IOP=ZJBOP(?). 

III.  o    is     equidistant 
from  A  and  B  (?). 

P  is  also  (?)  (219). 

.'.  OP  is  -L  to  AB  at  its  midpoint  (?)  (70). 

IV.  Arc  AX  =  arc  EX  (?)  (206). 


Q.E.D. 


221.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  equal  chords 
are  equally  distant  from  the  center. 

Given  :  O  O  ;  chord  AB  =  chord 
CD,  and  distances  OE  and  OF. 

To  Prove  :  OE  =  OF. 

Proof  :   Draw  radii  OA  and  OC. 
In  the    rt.   A  AOE   and    OOF, 
AE  =  \  AB  ;  CF  =  £  CD  (213). 
But  AB  =  CD  (Hyp.). 
Hence,  AE=  CF  (Ax.  3)  ;  and 


OE=  OF 


Q.E.D. 


222.   THEOREM.    In  the  same  circle  (or  in  equal  circles)  chords 
which  are  equally  distant  from  the  center  are  equal. 

Given  :  O  O  ;  chords  A  B  and  CD  ;  distance  OE=  distance  OF. 


BOOK  II 


89 


To  Prove :  chord  AB  —  chord  CD. 

Proof :  Draw  radii  OA  and  OC.  In  rt.  A  AOE  and  COF, 
AO  =  CO  (?);  OE  =  OF  (Hyp.).  .'.  A  AOE  =  A  COF  (?). 
.-.  AE  —  CF  (?).  AB  is  twice  AE  and  CD  is  twice  CF  (?). 

.'.  ^1J5  =  CD   (Ax.  3).  Q.E.D. 

223.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  if  two 
chords  are  unequal,  the  greater  chord  is  at  the  less  distance  from  the 
center. 

Given :  O  O ;  chord  AB  >  chord 
CZ>,  and  distances  OE  and  OF. 

To  Prove  :  OE  <  OF. 

Proof  :  Arc  AB  >  arc  CD  (?) 
(211,  I).  Suppose  arc  AH  taken 
on  arc  AB  =  arc  CD.  Draw  chord 
AH.  Draw  OK  J_  to  AH  cutting 
AB  at  I. 

Chord  AH  =  chord  CD  (?)  (210). 

Distance  OK  =  distance  OF  (?)  (221). 

But  OE  <  01  (?)  (77);  and  Ol  <  OK  (?)  (Ax.  5). 

.*.  O.E  <   OK  (Ax.   11).       .'.  OJ£  <   OF  (Ax.  6).  Q.E.D. 

224.  THEOREM.    In  the  same  circle  (or  in  equal  circles)  if  two 
chords  are  unequally  distant  from  the  center,  the  chord  at  the  less  dis- 
tance is  the  greater. 

Given  :  O  O ;  chords  AB  and  CD ;  distance  OE  <  distance  OF. 
To  Prove :  Chord  AB  >  chord  CD. 

Proof :  It  is  evident  that  chord  AB  <  chord  CD,  or  =  chord 
CD,  or  >  chord  CD.  Proceed  by  the  method  of  exclusion. 

Another  Proof :  On  OF  take  ox=  OE.  At  X  draw  a  chord 
jilS  JL  to  OX.  Ch.  BS  is  II  to  ch.  CD  (?).  .'.  arc  RS  >  arc 
CD  (Ax.  5).  .-.  ch.  ES  >  ch.  CD  (?). 

But  ch.  AB  =  ch.  RS  (?).     .-.  ch.  AB  >  ch.  CD  (Ax.  6). 

Q.E.D. 
225.   COR.    The  diameter  is  longer  than  any  other  chord. 


90 


PLANE   GEOMETRY 


226.  THEOREM.    Through  three  points,  not  in  the  same  straight 
line,  one  circumference  can  be  drawn,  and  only  one. 

Given  :   Points  A  and  B  and  C. 
To  Prove:    I.  (?).     II.  (?). 

Proof :  I.  Draw  lines  AB,  BC, 
AC.  Suppose  their  _L  bisectors,  OZ, 
OX,  OF,  be  drawn.  These  Js  will 
meet  at  a  point  (?)  (85).  Using 
O  as  center  and  OA,  O.B,  or  OC  as 
radius,  a  circumference  can  be 
described  through  A,  B,  C  (85). 

II.  These  Js  can  meet  at  only  one  point  (85)  ;  that  is, 
there  is  only  one  center.  The  distances  from  O  to  A,  O  to  B, 
O  to  (7,  are  all  equal  (85);  that  is,  there  is  only  one  radius. 
Therefore  there  is  only  one  circumference  (205).  Q.E.D. 

227.  COR.   A  circumference  can  be  drawn  through  the  vertices  of 
a  triangle,  and  only  one. 

228.  COR.    A  circumference  is  determined  by  three  points. 

229.  COR.   A  circumference  cannot  be  drawn  through  three  points 
which  are  in  the  same  straight  line.     [The  Js  would  be  II.] 

230.  COR.    A  straight  line  can  intersect  a  circumference  in  only 
two  points.  (229.) 

231.  COR.    Two  circumferences  can  intersect  in  only  two  points. 

232.  THEOREM.     If  two  circumferences  intersect,  the  line  joining 
their  centers  is  the  perpendicular  bisector  of  their  common  chord. 

Proof:  Draw  radii  in 
each  O  to  ends  of  AB. 
Point  O  is  equally  dis- 
tant from  A  and  B  (?). 
Point  C  is  equally  dis- 
tant from  A  and  B  (?). 
.*.  OCis  the  J_  bisector  of 

AB  (?)  (70).  Q.E.D. 


BOOK  II 


91 


233.   THEOREM.    Parallel   lines  intercept  equal  arcs  on  a  circum- 
ference. 

p  M  p 


Given :  A  circle  and  a  pair  of  parallels  intercepting  two 
arcs. 

To  Prove :  The  intercepted  arcs  are  equal. 
There  may  be  three  cases : 

I.  If  the  Us  are  a  tangent  (AB,  tangent  at  P)  and  a  secant 
(CD,  cutting  the  circle  at  E  and  F). 

Proof:  Draw  diameter  to  point  of  contact,  P.  This  di- 
ameter is  _L  to  AB  (216).  PP'  is  also  J_  to  EF  (?)  (95). 
.-.  arc  EP  =  arc  FP  (?)  (212). 

II.  If  the  Us  are  two  tangents  (points  of  contact  being  M 
and  N*). 

Proof :  Suppose  a  secant  be  drawn  II  to  one  of  the  tangents, 
cutting  O  at  E  and  S.  RS  will  be  II  to  other  tangent  (?)  (94). 

/,  arc  MR  =  arc  MS;  arc  RN  =  arc  SN  (proved  in  I). 
Adding,  arc  MRN=nrv  MSN  (Ax.  2). 

III.  If  the  Us  are  two  secants  (one  intersecting  the  O  at  A 
and  B;  the  other  at  C  and  D). 

Proof:  Suppose  a  tangent  be  drawn  touching  O  at  P,  II  to 
AB.  This  tangent  will  be  II  to  CD  (?). 

.*.  arc  PC=  arc  PD;  arc  PA  =  arc  PB  (by  I). 
Subtracting,  arc  AC  =  arc  BD  (Ax.  2).  Q.E.D. 


234.   A  polygon  is  inscribed 

in  a  circle,  or  a  circle  is  cir- 
cumscribed  about   a    polygon  J 


if  the  vertices  of  the  poly- 
gon are  in  the  circumference, 
and  its  sides  are  chords. 


92  PLANE   GEOMETRY 

A  polygon  is  circumscribed]    .f  ,,       .-,        .  ,-, 

J   .    .  it  the  sides  of  the  polygon 

about  a   circle,  or  a  circle  is  n  , 

are  all  tangent  to  the  circle. 
inscribed  in  a  polygon 

A  common  tangent  to  two  circles  is  a  line  tangent  to  both 
of  them. 

The  perimeter  of  a  figure  is  the  sum  of  all  its  bounding 
lines. 

EXERCISES  IN  DRAWING  CIRCLES 

1.  Draw  two  unequal  intersecting  circles.     Show  that  the  line  joining 
their  centers  is  less  than  the  sum  of  their  radii. 

2.  Draw  two  circles  externally  (not  tangent)  and  show  that  the  line 
joining  their  centers  is  greater  than  the  sum  of  their  radii. 

3.  Draw  two  circles  tangent  externally.     Discuss  these  lines  similarly. 

4.  Draw  two  circles  tangent  internally.     Discuss  these  lines  similarly. 

5.  Draw  two  circles  so  that  they  can  have  only  one  common  tangent. 

6.  Draw  two  circles  so  that  they  can  have  two  common  tangents. 

7.  Draw  two  circles  so  that  they  can  have  three  common  tangents. 

8.  Draw  two  circles  so  that  they  can  have  four  common  tangents. 

9.  Draw  two  circles  so  that  they  can  have  no  common  tangent. 

SUMMARY 

235.    The  following  summary  of  the  truths  relating  to  magnitudes, 
which  have  been  already  established   in  Book  II,  may  be  helpful  in 
attacking  the  original  work  following. 
I.  Arcs  are  equal  if  they  are : 

(1)  Intercepted  by  equal  central  angles. 

(2)  Subtended  by  equal  chords. 

(3)  Intercepted  by  parallel  lines. 

(4)  Halves  of  the  same  arc,  or  of  equal  arcs. 
II.  Lines  are  equal  if  they  are : 

(1)  Radii  of  the  same  or  equal  circles. 

(2)  Diameters  of  the  same  or  equal  circles. 

(3)  Chords  which  subtend  equal  arcs. 

(4)  Chords  which  are  equally  distant  from  the  center - 

(5)  Tangents  to  one  circle  from  the  same  point. 

III.   Unequal  arcs  and  unequal  chords  have  like  relations. 
[See  208;  211;  223;  224.] 


BOOK  II 


93 


ORIGINAL   EXERCISES 

1.  A  diameter  bisecting  a  chord  is  perpendicular  to  the 
chord  and  bisects  the  subtended  arcs.     [Use  70.] 

2.  A  diameter  bisecting  an  arc  is  the    perpendicular 
bisector  of  the  chord  of  the  arc.     [Draw  AR  and  BR.~\ 

3.  A  line  bisecting  a  chord  and  its  arc  is  perpendicular 
to  the  chord. 

4.  The  perpendicular  bisectors  of  the  sides  of  an  in- 
scribed polygon  meet  at  a  common  point. 

5.  A  line  joining  the  midpoints  of  two  parallel  chords 
passes  through  the  center  of  the  circle. 

[Supposediam.drawn_Lto^4J5;  this  will  be  _L  to  CD.  Etc.] 

6.  The  perpendiculars  to  the  sides  of  a  circumscribed 
polygon  at  the  points  of  contact  meet  at  a  common  point. 
[Use  217.] 

7.  The  bisector  of  the  angle  between  two  tangents  to  a  circle  passes 
through  the  center.     [Use  80.] 

8.  The  bisectors  of  the  angles  of  a  circumscribed  polygon  all  meet 
at  a  common  point. 

9.  Tangents  drawn  at  the  extremities  of  a  diameter  are  parallel. 

10.  In  the  figure  of  220,  prove  Z  APO  -Z.  ABO. 

11.  In  the  same  figure,  prove  ZPAB=ZPOB.  [Use  48.] 

12.  If  two  circles  are   concentric,   all  chords   of  the 
greater,  which  are  tangent  to  the  less,  are  equal. 

[Draw  radii  to  points  of  contact.     Use  216  ;  222.] 

13.  Prove  225  by  drawing  radii  to  the  ends  of  the  chord. 

14.  An  inscribed  trapezoid  is  isosceles.     [Use  233.] 

15.  The  line  joining  the  points  of  contact  of  two  parallel  tangents 
passes  through  the  center.    [Draw  radii  to  points  of  contact.     Etc.] 

16.  A  chord  is  parallel  to  the  tangent  at  the  midpoint 
of  its  subtended  arc.     [Draw  radii  to  point  of  contact 
and  to  the  ends  of  the  chord.     Also  draw  chords  of  the 
halves  of  the  given  arc.] 

17.  The  sum  of  one  pair  of  opposite  sides  of  a  cir- 
cumscribed quadrilateral  is  equal  to  the  sum  of  the 
other  pair.     [Use  219  four  times,  keeping  R  and  T  on 
the  same  side  of  the  equations.] 


94  PLANE   GEOMETRY 

18.  A  circumscribed  parallelogram  is  equilateral. 

19.  A  circumscribed  rectangle  is  a  square. 

20.  If  two  circles  are  concentric  and  a  secant  cuts  them  both,  the  por- 
tions of  the  secant  intercepted  between  the  circumferences   are   equal. 
[Use  212.] 

21.  Of  all  secants  that  can  be  drawn  to  a  circumference  from  a  fixed 
external  point,  the  longest  passes  through  the  center. 

To  Prove :  PB  >  PE. 

22.  The  shortest  line  from  an  external  point  to 
a  circumference  is  that  which,  if  produced,  would 
pass  through  the  center. 

To  Prove :  PA  <  PD.    Draw  CD. 

23.  If  two  equal  secants  be  drawn  to  a  circle  from  an  external  point, 
their  chord  segments  will  be  equal.      [Draw  OA , 

OP,  OC,   OB,   OD.     Prove    A  POD   and   POB 
equal;  then  &  COD  and  A  OB  are  equal.] 

24.  In  No.  23  prove  the  external  segments 
equal. 

25.  State  and  prove  the  converse  of  No.  23. 

26.  If  two  equal  secants  be  drawn  to  a  circle  from  an  external  point, 
they  will  be  equally  distant  from  the  center. 

27.  If  two  equal  chords  intersect  on  the  circumference,  the  radius 
drawn  to  their  point  of  intersection  bisects  their  angle. 

[Draw  radii  to  the  other  extremities  of  the  chords.] 

28.  Any  two  parallel  chords  drawn  through  the  ends  of  a  diameter 
are  equal. 

29.  If  a  circle  be  inscribed  in  a  right  triangle, 
the  sum  of  the  diameter  and  hypotenuse  will  be 
equal  to  the  sum  of  the  legs. 

[Draw  radii  OR,  OS',  ROSC  is  a  square  (?) ; 
then  prove  diameter  +  AB  =  A  C  +  BC.~] 

30.  Of  all  chords  that  can  be  drawn  through  a  given  point  within  a 
circle,  the  chord  perpendicular  to  the  diameter  through 

the  given  point  is  the  shortest. 

Given:   P,  the  point;  BOG  the  diam.;    LS  _L  to  BC    ° 
at  P;    GR   any  other  chord  through  P.     To  Prove:  (?). 

Proof:  Draw  OA  ±  to  GR.    Etc. 


BOOK   II 


95 


31.  What  is  the  longest  chord  that  can  be  drawn  through  a  given  point 
within  a  circle? 

32.  If  the  line  joining  the  point  of  intersection  of  two  chords  and 
center  bisects  the  angle  formed  by  the  chords,  they  are 

equal.     [Draw  Js  OE  and  OF  and  prove  them  = .     Etc.] 

33.  AB  and  AC  are  two  tangents  from  A  ;  in  the  less 
arc  EC  a  point  D  is  taken  and  a  tangent  drawn    at  Z>, 
meeting  AB  at  E  and  AC  at  F]  AE  +  EF  +  .4Fequals  a 
constant  for  all  positions  of  D  in  arc  BC. 

[Prove  this  sum  =  AB  +  A C.] 

34.  The  radius  of  the   circle   inscribed  in   an  equi- 
lateral triangle  is  half  the  radius  of  circle  circumscribed 
about  it.     [Use  152.] 

35.  If  the  inscribed  and   circumscribed  circles  of  a 
triangle  are  concentric,  the  triangle  is  equilateral.  jj^ 

36.  If  two  parallel  tangents  meet  a  third  tangent 
and  lines  be  drawn  from   the  points  of  intersection 
to  the  center,  they  will  be  perpendicular. 

37.  Tangents  drawn  to  two  tangent  circles  from 
any  point  in  their  common  interior  tangent  are  equal. 

38.  The  common  interior  tangent  of  two  tangent 
circles  bisects  their  common  exterior  tangent. 

39.  Do  the  theorems  of  No.  37  and  No.  38  apply 
if  the  circles  are  tangent  internally  ?    If  so,  prove; 

40.  In  the  adjoining  figure  if   AE  and  AD  are 
secants,  A  E  passing  through  the  center,  and  the  ex- 
ternal  part   of  AD  is  equal  to  a  radius,   the  angle 
DCE  =  3ZA. 

[Draw  BC.    Z  DEC  =  ext.  /.  of  A  ABC  = 
=  Z  D  (explain).     Z  DCE  =  an  ext.  Z,  etc.] 

41.  If  perpendiculars  be  drawn  upon  a  tangent 
from  the  ends  of  any  diameter : 

(1)  The  point  of  tangency  will  bisect  the  line 
between  the  feet    of  the    perpendiculars. 

[Draw  CP.     Use  144.]  ~~O P         e~~ 

(2)  The  sum  of  the  perpendiculars  will  equal  the  diameter. 

(  3)  The  center  will  be  equally  distant  from  the  feet  of  the  perpendic- 
ulars.    [Use  67.] 


96  PLANE   GEOMETRY 

42.    The  two  common  interior  tangents  of  two  circles  are  equal. 
A 


C 

43.  The  common  exterior  tangents  to  two  circles  are  equal. 
[Produce  them  to  intersection.] 

44.  In  the  above  figure,  prove  that  RH  =  SF. 

Proof:  AR  +  RB  =  CS  +  SD  ;  .'.  AR  +  (RH  +  HF)  =  (SF  + 
HF)  +  SD. 

.'.  RH  +  RH  +  HF  =  SF  +  HF  +  SF ;  /.  2  RH  =  2  SF,  etc. 
Give  reasons  and  explain. 

45.  The  common  exterior  tangents  to  two  circles  intercept  on  a  com- 
mon interior  tangent  (produced),  a  line  equal  to   a   common  exterior 
tangent.     To  Prove  :    RS  =  AB. 

46.  Prove  that  in  the  figure  of  No.  42  the  line  joining  the  centers 
will  contain  0  and  0'. 

47.  Prove  that  in  .the  figure  of  No.  42  if   chords  A  C  and  BD  are 
drawn,  they  are  parallel. 

48.  If  a  circle  be  described  upon  the  hypotenuse  of  a  right  triangle  as 
a  diameter,  it  will  contain  the  vertex  of  the  right  angle  (148). 

49.  The  median  of  a  trapezoid  circumscribed  about  a  circle  equals 
one  fourth  the  perimeter  of  the  trapezoid. 

50.  If  the  extremities  of  two  perpendicular  diameters  be  joined  (in 
order),  the  quadrilateral  thus  formed  will  be  a  square. 

51.  If  any  number  of  parallel  chords  of  a  circle  be  drawn,  their  mid- 
points will  be  in  the  same  straight  line. 

52.  State  and  prove  the  converse  of  No.  35. 

53.  The  line  joining  the  center  of  a  circle  to  the  point 
of  intersection  of  two  equal  chords  bisects  the  angle  formed 
by  the  chords. 


BOOK  II  97 

KINDS  OF   QUANTITIES  — MEASUREMENT 

236.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another  —  both  being  of  the  same  kind. 

237.  To  measure  a  quantity  is  to  find  the  number  of  times 
it   contains  another  quantity  of  the  same  kind,    called  the 
unit.     This  number  is  the  ratio  of  the  quantity  to  the  unit. 

238.  Two  quantities  are   called   commensurable   if   there 
exists  a  common  unit  of  measure  which  is  contained  in  each 
a  whole  (integral)  number  of  times. 

Two  quantities  are  called  incommensurable  if  there  does 
not  exist  a  common  unit  of  measure  which  is  contained  in 
each  a  whole  number  of  times. 

Thus  :   $17  and  $  35  are  commensurable,  but  $  17  and  $  \/35  are  not. 
Two  lines  18£  ft.  and  13  yd.  are  commensurable,  but  18£  in.  and 
Vl3  mi.  are  not. 

239.  A  constant  quantity  is  a  quantity  whose  value  does 
not  change  (during  a  discussion).      A  constant  may  have 
only  one  value. 

A  variable  is  a  quantity  whose  value  is  changing.  A 
variable  may  have  an  unlimited  number  of  values. 

240.  The  limit  of  a  variable  is  a  constant,  to  which  the 
variable  cannot  be  equal,  but  from  which  the  variable   can 
be  made  to  differ  by  less  than  any  mentionable  quantity. 

241.  Illustrative.     The  ratio  of  15  yd.  to  25  yd.  is  written  either  |f 
or  15  -4-  25  and  is  equal  to  three  fifths.     If  we  state  that  a  son  is  two 
thirds  as  old  as  his  father,  we  mean  that  the  son's  age  divided  by  the 
father's  equals  two  thirds.     A  ratio  is  a  fraction. 

The  statement  that  a  certain  distance  is  400  yd.  signifies  that  the  unit 
(the  yard),  if  applied  to  this  distance,  will  be  contained  exactly  400  times. 

Are  $7.50  and  $3.58  commensurable  if  the  unit  is  $1?  Idime? 
1  cent? 

Are  10  ft.  and  Vl9  ft.  commensurable? 


98  PLANE   GEOMETRY 

The  height  of  a  steeple  is  a  constant ;  the  length  of  its  shadow  made 
by  the  sun  is  a  variable.  Our  ages  are  variables.  The  length  of  a 
standard  yard,  mile,  or  meter,  etc.,  is  a  constant.  The  height  of  a  grow- 
ing plant  or  child  is  a  variable. 

The  limit  of  a  variable  may  be  illustrated  by  considering  a  right  tri- 
angle ABC,  and  supposing  the  vertex  A  B 
to  move  farther  and  farther  from  the 
vertex  of  the  right  angle.     It  is  evident 
that  the  hypotenuse  will  become  longer, 
that  AC  will  increase,  but  EC  will  re- 
main the  same  length.     The  angle  A 
must   decrease,  the  angle  B  must  in- 
crease, but  the   angle  C  remains  con-        A  * —      A  « —  A 
stantly  a  right  angle.     If  we  carry  vertex  A  toward  the  left  indefinitely, 
the  Z  A  will  become  less  and  less  but  cannot  become  zero.     [Because, 
then  there  could  be  no  A.] 

Hence,  the  limit  of  the  decreasing  /.  A  is  zero  (240). 

Likewise,  the  Z.  B  will  become  larger  and  larger  but  cannot  become 
equal  to  a  right  angle.  [Because,  then  two  sides  of  the  triangle  would  be 
parallel,  which  is  impossible.]  But  it  may  be  made  as  nearly  equal  to  a 
right  angle  as  we  choose. 

Hence,  the  limit  of  /.  B  is  a  right  angle  (240). 

To  these  limits  we  cannot  make  the  variables  equal,  but  from  these 
limits  we  can  make  them  differ  by  less  than  any  mentionable  angle,  how- 
ever small. 

The  following  supplies  another  illustration  of  the  limit  of  a  variable. 
The  sum  of  the  series  l+4  +  |  +  |  + T*  +  *V  +  &  +  T£*  +  e*c.  etc.,  will 
always  be  less  than  2,  no  matter  how  many  terms  are  collected.  But  by 
taking  more  and  more  terms  we  can  make  the  actual  difference  between 
this  sum  and  2  less  than  any  conceivable  fraction,  however  small. 
Hence,  2  is  the  limit  of  the  sum  of  the  series.  The  limit  is  not  3  nor  4, 
because  the  difference  between  the  sum  and  3  cannot  be  made  less  than 
any  assigned  fraction.  Neither  is  the  limit  1^.  (Why  not?)  Similarly, 
the  limit  of  the  value  of  .333333 ad  infinitum  is  £. 

Certain  variables  actually  become  equal  to  a  fixed  magnitude ;  but 
this  fixed  magnitude  is  not  a  limit  (240).  Thus  the  length  of  the 
shadow  of  a  tower  really  becomes  equal  to  a  fixed  distance  (at  noon). 
A  man's  age  really  attains  to  a  definite  number  of  years  and  then  ceases 
to  vary  (at  death). 

Such  variables  have  no  limit  in  the  mathematical  sense  of  that 
word. 


BOOK  II  99 

Hence  :  If  a  variable  approaches  a  constant,  and  the 
difference  between  the  two  can  be  made  indefinitely  small 
but  they  cannot  become  equal,  the  constant  is  the  limit  of 
the  variable.  This  is  merely  another  definition  of  a  limit. 

242.  THEOREM  OF  LIMITS.    If  two  variables  are  always  equal  and 
each  approaches  a  limit,  their  limits  are  equal. 

Given:  Two  variables  v  and  v1 ';  v  always  =v'-9  also  v  ap- 
proaching the  limit  I;  v1  approaching  the  limit  V . 

To  Prove :  l—ll. 

Proof:  v  is  always  =  vf  (Hyp.).  Hence  they  may  be  con- 
sidered as  a  single  variable.  Now  a  single  variable  can 
approach  only  one  limit  (240).  Hence,  1=  I' .  Q.E.D. 

NOTE.  In  order  to  make  use  of  this  theorem,  one  must  have,  jirst,  two 
variables;  second,  these  must  be  always  equal;  third,  they  must  each  ap- 
proach a  limit.  Then,  the  limits  are  equal. 

243.  (1)   Algebraic  principles  concerning  variables. 
If  v  is  a  variable  and  k  is  a  constant : 

I.    v  4-  k  is  a  variable.  IV.    kv  is  a  variable. 

II.    v  —  k  is  a  variable.  V.    -  is  a  variable. 

k 

III.    k  ±  v  is  a  variable.  VI.    -  is  a  variable. 

v 

These  six  statements  are  obvious. 

(2)   Algebraic  principles  concerning  limits. 

If  v  is  a  variable  whose  limit  is  ?,  and  k  is  a  constant : 
I.    v  ±  Jc  will  approach  I  ±  k  as  a  limit. 
II.    k  ±  v  will  approach  k  ±  I  as  a  limit. 

III.  kv  will  approach  kl  as  a  limit. 

IV.  ^  will  approach  -  as  a  limit. 

K  K 

k  k 

V.  -  will  approach  -  as  a  limit. 
v  I 

NOTE.  In  these  principles  as  applied  to  Plane  Geometry,  a  variable 
is  not  added  to,  nor  subtracted  from,  nor  multiplied  by,  nor  divided  by 
another  variable.  These  operations  present  little  difficulty,  however. 


100  PLANE   GEOMETRY 

Proofs :  I.   v  cannot  =  /  (240).     .-.  v±k  cannot  =  I  ±  k. 

Also,  v  —  I  approaches  zero  (240). 

.•.  (v  ±  k)  —  (I  ±  k)  approaches  zero.     (Because  it  reduces  to  v  —  I.) 

Hence,  v  ±  k  approaches  /  ±  k  (240). 

II.   Demonstrated  similarly. 

III.   If  kv  =  kl,  then  v  =  I  (Ax.  3).    But  this  is  impossible  (240). 
.-.  kv  cannot  =  kl. 

Also  v  —  I  approaches  zero  (240). 
.•.  k(v  —  1)  or  kv  —  kl  approaches  zero. 
Therefore  kv  approaches  kl  (240). 
IV  and  V.   Demonstrated  similarly. 

244.   THEOREM.    In  the  same  circle  (or  in  equal  circles)  the  ratio 
of  two  central  angles  is  equal  to  the  ratio  of  their  intercepted  arcs. 


Y 
Given  :  O  o  =  O  C  ;  central  A  O  and  C;  arcs  AB  and  XY. 

To  Prove  :   =  ^-2 

Z.  C      arc  XY 

Proof :  I.  If  the  arcs  are  commensurable.  There  exists  a 
common  unit  of  measure  of  AB  and  XY  (238).  Suppose 
this  unit,  when  applied  to  the  arcs,  is  contained  5  times  in 

AB  and  7  times  in  XY.     .-.  arc  AB  -  g  (Ax.  3).     Draw  radii 

to    the    several    points  of    division    of   the   arcs.      Z  O   is 
divided  into  5  parts,  Z  c  into  7  parts  ;  all  of  these  twelve 

parts  are  equal  (?)  (207).     .-.—  =  |  (Ax.  3). 

/_  c      i 

Z  O  _  arc  A  B   f  »       .,  ^ 
''"7~C~l^^^  >'  Q.K.D. 


BOOK  II  101 

II.  If  the  arcs  are  incommensurable.  There  does  not 
exist  a  common  unit  (238).  Suppose  arc  AB  divided  into 
equal  parts  (any  number  of  them).  Apply  one  of  these  as 
a  unit  of  measure  to  arc  XT.  There  will  be  a  remainder 
PFleft  over.  (Because  AB  and  XT  are  incommensurable.) 


Draw  CP.     Now    Z°    =  arc  AB.    (The  case  of  commen- 
Z  XCP     arc  XP 

surable  arcs.) 

Indefinitely  increase   the  number   of  subdivisions  of  arc 
AB.     Then  each  part,  that  is,  our  unit  or  divisor,  will  be 
indefinitely  decreased.     Hence  PF,  the  remainder,  will  be  in- 
definitely decreased.    (Because  the  remainder  <  the  divisor.) 
That  is,  arc  PF  will  approach  zero  as  a  limit 
and  Z  PCF  will  approach  zero  as  a  limit. 

.'.  arc  XP  will  approach  arc  XT  as  a  limit  (240) 
and  Z  XCP  will  approach  Z  XCT  as  a  limit  (240). 

Z  °      will  approach  -^-°—  as  a  limit  (243) 


Z  XCP  Z  XCT 

j  arc  AB      -n  T    arc  AB  v     •  , 

and  will  approach  -         -  as  a  limit 

arc  XP  arc  XT 

Z  o        arc  AB  (?)  (242) 


Z  XCF     arc  XF 


Ex.  How  many  degrees  are  there  in  a  central  angle  which  intercepts 
£  of  the  circumference?  £  of  the  circumference?  ^  of  the  circumfer- 
ence? of  the  circumference? 


102 


PLANE   GEOMETRY 


245.   THEOREM.    A  central  anglers  measured  by  its  intercepted  arc. 

Given:    O  0;  /.  AOY \  arc  AY. 

To  Prove:  Z  AOT  is  measured 
by  the  arc  AY,  that  is,  they  contain 
the  same  number  of  units. 

Proof :  The  sum  of  all  the  A 
about  0=4  rt.  Zs=  360°  (?)  (47). 

If  the  circumference  of  this  O 
be  divided  into  360  equal  parts  and 
radii  be  drawn  to  the  several  points 
of  division,  there  will  be  360  equal  central  A  (207). 

Each  of  these  360  central  angles  will  be  a  degree  of  angle  (21). 

Suppose  we  call  each  of  the  360  equal  arcs,  a  degree  of  are. 
Take  Z  AOT,  one  of  these  degrees  of  angle,  and  arc  AT,  one 


of  the  degrees  of  arc.     Then, 


Z^tOF^arc  AY 
Z  AOT      arc  AT 


(?)  (244). 


But =  Z4OF-*-a  unit  of   angle  =  the   number  of 

Z  AOT 

units  in  2.AOY  (237). 

And =  arc   AY  -4-  a   unit   of  arc  =  the   number  of 

arc  AT 

units  in  arc  AY  (237). 

Hence,  the  number  of  units  in  Z  AOY=  the  number  of  units 
in  arc  AY  (Ax.  1). 

That  is,  Z  AOY  is  measured  by  arc  AY.  Q.E.D. 

246.  COR.    A   central  right  angle   intercepts  a  quadrant   of  arc. 
(Because  each  contains  90  units.) 

"/C  ^s. 

247.  COR.   A  right  angle  is  measured  by  half 
a  semicircumf  erence,  that  is,  by  a  quadrant. 

248.  An  angle  is  inscribed  in  a  segment 
if  its  vertex  is  on  the  arc  and  its  sides  are 
drawn  to  the  ends  of  the  arc  of  the  segment. 

Thus  ABCD  is  a  segment  and  Z  ABD  is  inscribed  in  it. 


BOOK  II 


103 


249.  THEOREM.  An  inscribed  angle  is  measured  by  half  its  inter- 
cepted arc. 

Given  :    O  O  ;  inscribed  Z  A  ;  arc  CD. 

To  Prove :    Z  A  is  measured  by  J  arc  CD. 

Proof:  I.  If  one  side  of  the  Z  is  a 
diameter.  Draw  radius  CO.  A  ^.OC  is 
isosceles  (?).  .-.  ZA  =  ZC  (?). 

Z  COD  =  ZA  +  Z  C  (?)  (108). 

/.  Z  COD  =  Z^l+Z^  =  2Zj.  (Ax.  6). 

That  is,  Z  A  =  £  Z  COD  (Ax.  3). 

But  Z  COD  is  measured  by  arc  CD  (?)  (245). 

.*.  \  Z  COD  is  measured  by  J  arc  CD  (Ax.  3). 

Therefore,  Z  A  is  measured  by  J  arc  CD  (Ax.  6). 


II.  If  the  center  is  within  the  angle.     Draw  diameter  AX. 
Z.CAX  is  measured  by  J  arc  CX  (I). 

is  measured  by  J  arc  DX  (I).    Adding, 

is  measured  by  ^  arc  CD  (Ax.  2). 

III.  If  the  center  is  without  the  angle.    Draw  diameter  AX. 

is  measured  by  J  arc  CX  (I). 
T  is  measured  by  J  arc  DX  (I).     Subtracting, 

Z  C^ID  is  measured  by  J  arc  CD    (Ax.  2).  Q.E.D. 

NOTE.     It  is  evident  that  angles  measured  by  $•  the  same  arc  are  equal. 


are 


Ex.  1.   In  the  figure  of  249,  if  arc  CD  is  56°,  how  many  degrees  — 
there  in  angle  A  ?    If  arc  CD  is  108°,  how  many  degrees  are  in  angle  A  ? 
Ex.  2,   If  ^  A  contains  35°,  how  many  degrees  are  there  in  arc  CZ>? 


104 


PLANE   GEOMETRY 


250.  THEOREM.  All  angles  inscribed  in  the 
same  segment  are  equal. 

Given  :   The  several  A  A  inscribed  in    A 
segment  BAC. 

To  Prove :  These  angles  all  equal. 

Proof :  Each  Z  BAC  is  measured  by 
1  arc  BC  (?)  (249). 

.-.  they  are  equal.  (Because  they  are 
measured  by  half  the  same  arc.)  Q.E.D. 


251.   COR.   All  angles  inscribed  in  a  semi- 
circle are  right  angles. 

Proof  :    Each  is  measured  by  half  of  a 
semicircumference  (?)  (249). 

/.  each  is  a  rt.  Z  (?)  (24T).         Q.E.D. 


252.   THEOREM.   The  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  half  the  intercepted  arc. 

Given  :  Tangent  TN ; 
chord  PA  ;  Z  TPA ;  arc 
PA. 

To  Prove  :  Z  TPA  is 
measured  by  ^  arc  PA. 

Proof  :    Draw   diameter 

PX  to    point    of    contact. 

Z  TPX  is  a  rt.  Z  (?)  (216)  ; 

arc  PAX  is  a  semicircum- 

ference  (?)   (204). 

Z  TPX  is  measured  by  J  arc  PAX  (?)  (247). 

Z  APX  is  measured  by  -|-  arc  AX  (?)  (249).  Subtracting, 

Z  TPA  is  measured  by  ^  arc     PA    (Ax.  2).  Q.E.D 

Similarly,  Z  NPA  is  measured  by  %  arc  PBA. 

(Use  Z  -flTP-*,  and  add.) 


BOOK  II  105 


253.  THEOREM.   The    angle  formed   by   two   chords    intersecting 
within  the  circumference  is  measured  by  half  the  sum  of  the  inter 
cepted  arcs.     (The     arcs    are    those 

intercepted    by   the    given    angle 
and  by  its  vertical  angle.) 

Given  :  Chords  AB  and  CD  in- 
tersecting at  P;  Z  APC\  arcs  AC 
and  DB. 

To  Prove  :  Z  A  PC  is  measured 
by  ^  (arc  AC+  arc  DB). 

Proof :  Suppose  CX  drawn  through  C  \\  to  AB. 

Now  Z  C  is  measured  by  J  arc  DX  (?)  (249). 

That  is,  Z  C  is  measured  by  J  (arc  BX+  arc 

But  Z  C  =  Z  ^LPC  (?)  (97)  and  arc  J?x  =  arc  -4C(?)  (233). 

.*.  Z.APC  is  measured  by  |  (arc  -4(7  +  arc  DB)  (?)  (Ax.  6). 

Q.E.D. 

NOTE.     This  theorem  may  be  proved   by  drawing  chord  AD.     Then 
ZAPC  is  an  ext.  ^  of  &ADP  and  =  /.A  +  Zi>  (?).     [Use  249.] 

254.  THEOREM.  The  angle  formed  by  two  tangents  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 

Given :  The  two  tan- 
gents AC  and  AB;  Z.A\ 
arcs  CMB  and  CNB. 

To  Prove  : 

Z  A  is  measured  by 
|(arc  O¥B  —  arc  CNB). 

Proof :  Suppose  CX 
drawn  ||  to  AB. 

Now  Z  DCJT  is  measured  by  J  arc  CX  (?)  (252). 
That  is,  Z  DCX  is  measured  by  J  (arc  CMB  —  arc 
But  Z  DCX  =Z  ^1  (?)  (98)  ;   arc  BX  =  arc  CNB  (?). 
/.Z.4  is  measured  by  ^  (arc  CMB  —  arc  CJVB)  (Ax. 6).  Q.E.D. 


106 


PLANE   GEOMETRY 


255.  THEOREM.  The  angle  formed  by  two  secants  which  intersect 
without  the  circumference  is  measured  by  half  the  difference  of  the  in- 
tercepted arcs. 

Given:   (?).     To  Prove:  (?). 
Proof:    Suppose   BX  drawn. 

Where?  How? 

Z  CBX  is  measured  by 
1  arc  CX  (?). 

That  is,  by  J(arc  CE 
—  arc  EX) . 

But  Z  CBX  =  ZA  (?);  arc  #JT=arc  BD  (?). 

.'.  ZA  is  meas.  by  J  (arc  CJ£— arc^D)  (Ax.  6).        Q.E.D. 

256.  THEOREM.  The  angle  formed  by  a  tangent  and  a  secant  which 
intersect  without  the  circumference,  is  measured  by  half  the  difference 
of  the  intercepted  arcs.  c 

Given:  (?). 

To  Prove :  (?). 

Proof:   Suppose  BX  drawn 
etc. 

Z  CBX  is  measured  by  J 


?)  (252). 

That  is,  by^  (arc  BXE—    A 
arc  EX).     Etc. 


NOTE.     The  theorem  of  254  may  be  proved  by  drawing  chord  BC. 
Then  Z  DCB  =  Z  A  +  Z  CBA  (?)  ;  or,  Z  A  =  Z  DCB  -  /.  CBA  (Ax.  2). 
Z  DOB  is  measured  by  \  arc  CMB  (?)  and  Z  CBA  is  measured  by  £ 
(?).     Hence,  Z  A  is  measured  by  $  (arc  CMB  -  arc  CW5)  (?). 


Ex.  1.   Prove  the  theorem  of  253  for  angle  A  PD  by  drawing  chord  A  C. 

Ex.  2.  Prove  the  theorem  of  255  by  drawing  chord  CD.  Again,  by 
drawing  chord  BE. 

Ex.  3.  Prove  the  theorem  of  256  by  drawing  chord  BD.  Again,  by 
chord  BE. 


BOOK  II  107 

ORIGINAL   EXERCISES 

1.  If  an  inscribed  angle  contains  20°,  how  many  degrees  are  there  in 
its  intercepted  arc  ?     How  many  degrees  are  there  in  the  central  angle 
which  intercepts  the  same  arc  ? 

2.  A  chord  subtends  an  arc  of  74°.     How  many  degrees  are  there 
in  the  angle  between  the  chord  and  a  tangent  at  one  of  its  ends? 

3.  How  many  degrees  are  there  in  an  angle  inscribed  in  a  segment 
whose  arc  contains  210°?  in  a  segment  whose  arc  contains  110°?   40°? 

4.  Two  intersecting  chords  intercept  opposite  arcs  of  28°  and  80°. 
How  many  degrees  are  there  in  the  angle  formed  by  the  chords  ? 

5.  The  angle  between  a  tangent  and  a  chord  contains  27°.  How  many 
degrees  are  there  in  the  intercepted  arc? 

6.  The  angle  between  two  chords  is  30°;  one  of  the  arcs  intercepted 
is  40° ;  find  the  other  arc.     [Denote  the  arc  by  x.~\ 

7.  If  in   figure  of  252,  arc  AP  contains  124°,    how  many  degrees 
are  there  in  Z  APX  ?    in  Z  NPA  ? 

8.  If  in  figure  of  253,  arc  AC  is  85°,  Z  APC  is  47°,  find  arc  DB. 

9.  If  the  arcs  intercepted  by  two  tangents  contain  80°  and  280°,  find 
the  angle  formed  by  the  tangents. 

10.  If  the  arcs  intercepted  by  two  secants  contain  35°  and  185°,  find 
the  angle  formed  by  the  secants. 

11.  If  in  figure  of  254,  arc  CB  is  135°,  find  the  angle  A. 

12.  If  in  figure  of  255,  angle  A  =  42°  and  arc  BD  =  70°,  find  arc  CE. 

13.  If  in  figure  of  256,  angle  A  =  18°,  arc  BXE  =  190°,  find  arc  BD. 

14.  If  the  angle  between  two  tangents  is  80°,   find  the  number  of 
degrees  in  each  intercepted  arc.     [Denote  the  arcs  by  a;  and  360° — x.~\ 

15.  The  circumference  of   a  circle  is  divided  into  four  arcs,  70°,  80°. 
130°,  and  x.     Find  x  and  the  angles  of  the  quadrilateral  formed  by  the 
chords  of  these  arcs. 

16.  Find  the  angles  formed  by  the  diagonals  in  quadrilateral  of  No.  15. 

17.  Three  of  the  intercepted  arcs  of  a  circumscribed  quadrilateral  are 
68°,  98°,  114°.     Find  the  angles  of  the  quadrilateral.     If  the  chords  are 
drawn  connecting  (in  order)  the  four  points  of  contact,  find  the  angles  of 
this  inscribed  quadrilateral.    Also  find  the  angles  between  its  diagonals. 


108 


PLANE   GEOMETRY 


18.  If  the  angle  between  two  tangents  to  a  circle  is  40°,  find  the  other 
angles  of  the  triangle  formed  by  drawing  the  chord  joining  the  points  of 
contact. 

19.  The  circumference  of  a  circle  is  divided 
into  four   arcs,  three  of  which   are,   RS  =  62°, 
ST  =  142°,  TU  =  98°.     Find : 

(1)  Arc  UR', 

(2)  The  three  angles  at  R-,  at  S]  at  T7;  at  U; 

(3)  The  angles  A,  B,  C,  D  of  circumscribed 
quadrilateral ; 

(4)  The  angles  between  the  diagonals  RT  and  SU; 

(5)  The  angle  between  RU  and  ST  at  their  point  of  intersection 
(if  produced) ; 

(6)  The  angle  between  RS  and  TU  at  their  intersection; 

(7)  The  angle  between  AD  and  BC  at  their  intersection ; 

(8)  The  angle  between  AB  and  DC  at  their  intersection; 

(9)  The  angle  between  RS  and  DC  at  their  intersection; 
(10)  The  angle  between  AD  and  ST  at  their  intersection. 

20.  If  in  the  figure  of  No.  19,  Z  A  =  96° ;  Z  B  =  112° ;  and  Z  C  =  68°, 
find  the  angles  of  the  quadrilateral  RSTU.      [Denote  arc  RU  by  x. 

257.   It  is  evident  from  the  theorems  relating  to  the  measurement  of 
angles,  that : 

1    Equal  angles  are  measured  by  equal  arcs  (in  the  same  circle) . 
2.  Equal  arcs  measure  equal  angles.  T- 

21.  Prove  theorem  of  252  by  drawing  chord  parallel 
to  the  tangent. 

22.  The  opposite  angles  of  an  inscribed  quadri- 
lateral are  supplementary. 

Proof :  ^  A  +  Z  C  are  meas.  by  1  circumference. 

23.  If  two  chords   intersect  within  a  circle,  and   at 
right  angles,  the  sum  of  one  pair  of  opposite  arcs  equals 
the  sum  of  the  other  pair.  [Use  253.] 

24.  If  a  tangent  and  a  chord  are  parallel,  and  the  chords  of  the  two 
intercepted  arcs  be  drawn    they  will  make  equal  angles  with  the  tan- 
gent.    [Use  233 ;  252.] 

25.  The  line  bisecting  an  inscribed  angle  bisects  the  intercepted  arc. 


BOOK  II 


109 


26.  The  line  joining  the  vertex  of  an  inscribed  angle  to  the  midpoint 
of  its  intercepted  arc  bisects  the  angle.  T  p 

27.  The  line  bisecting  the  angle  between  a  tangent 
and  a  chord  bisects  the  intercepted  arc. 

28.  State  and  prove  the  converse  of  No.  27. 

29.  The  angle  between   a  tangent  and  a  chord  is  half  the  angle 
between  the  radii  drawn  to  the  ends  of  the  chord. 

30.  If  a  triangle  be  inscribed  in  a  circle  and  a  tan- 
gent be  drawn  at  one  of  the  vertices,  the  angles  formed 
between  the  tangent  and   the   sides  will  equal  the 
other  two  angles  of  the  triangle. 

31.  By  the  figure  of  No.  30  prove  that  the  sum  of  the  three  angles  of 
a  triangle  equals  two  right  angles. 

32.  If  one  pair  of  opposite  sides  of  an  inscribed  quad- 
rilateral are  equal,  the  other  pair  are  parallel. 

Proof :  Draw  Js  BX,  CY;  arc  AB  =  arc  CD  (?). 

.-.  arc  A BC  =  arc  BCD  (Ax.  2) .  A\x y/D 

Hence  prove  rt.  &  ABX  and  CDY  equal. 

33.  If  any  pair  of  diameters  be  drawn,  the  lines  joining  their  extremi- 
ties (in  order)  will  form  a  rectangle.     [Use  251.] 

34.  If  two  circles  are  tangent  externally  and 
any  line  through  their  point  of  contact  intersects 
the  circumferences  at  B  and  C,  the  tangents  at  B 
and  C  are  parallel.     [Draw  common  tangent  at  A. 
Prove  :  Z.  A  CT  =  Z  ABS.] 

35.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

36.  If  two  circles  are  tangent  externally  and  any  line  be  drawn  through 
their  point  of  contact  terminating  in  the  circumferences,  the  two  diam- 
eters drawn  to  the  extremities  will  be  parallel. 

37.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

38.  If  two  circles  are  tangent  externally  and 
any  two   lines   be  drawn   through  their  point  of 
contact  intersecting  their  circumferences,  the  chords 
joining  these  points  of  intersection  will  be  parallel. 

[Draw  common  tangent  at  0.    Prove :  Z  C  =  Z  />.] 

39.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 


110 


PLANE   GEOMETRY 


40.  The  circle  described  on  one  of  the  equal  sides  of  an  isosceles 
triangle  as  a  diameter  bisects  the  base. 

Proof:  Draw  line  BM.     The  A  are  rt.  &  (?)  and  equal  (?). 

41.  If  the  circle,  described  on  a  side  of  a  triangle,  as 
diameter,  bisects  another  side,  the  triangle  is  isosceles. 

42.  All  angles  that  are  inscribed  in  a  segment  greater 
than  a  semicircle  are  acute,  and  all  angles  inscribed  in  a 
segment  less  than  a  semicircle  are  obtuse. 

43.  An  inscribed  parallelogram  is  a  rectangle. 
Proof:  Arc  AB  =  arc  CD  (?)  ;    arc  BC  =  arc  AD  (?). 
.'.  arc  ABC  =  arc  ADC  (?) ;  that  is,  each  =  a  semicir- 

cumference.     Etc. 

44.  The  diagonals  of  an  inscribed  rectangle  pass  through  the  center 
and  are  diameters.  e 

45.  The  bisectors  of  all  the  angles  inscribed  in  the 
same  segment  pass  through  a  common  point. 

46.  The  tangents   at  the  vertices  of  an  inscribed 
rectangle  form  a  rhombus. 

[A  A  BF  and  HD  C  are  isosceles  (?)  and  equal  ?    Etc.] 

47.  If   a   parallelogram  be  circumscribed   about   a 
circle,  the  chords  joining  (in  order)  the  four  points  of 
contact  will  form  a  rectangle.     [Prove  BD  a  diameter.] 

48.  A  circumference  described  on  the  hypotenuse  of 

a  right  triangle  as  a  diameter  passes.through  the  vertices  A* 
of  all  the  right  triangles  having  the  same  hypotenuse. 

49.  If  from  one  end  of  a  diameter  a  chord  be  drawn,  a  perpendicular  to 
it  drawn  from  the  other  end  of  the  diameter  will  inter- 
sect the  first  chord  on  the  circumference.     [Use  148.] 

50.  If  two  circles  intersect  and  a  diameter  be  drawn 

in  each  circle  through  one  of  the  points  of  intersection,  .Cv 

the  line  joining  the  ends  of  these  diameters  will  pass 

through  the  other  point  of  intersection.    [Draw  chord  AB.    Use  251 ;  43.] 

51.  If  ABCD  is  an  inscribed  quadrilateral,  AB  and  DC  produced  to 
meet  at  E,  AD  and  BC  produced  to  meet  at  F,  the  bisectors  of  angles 
E  and  F  are  perpendicular. 

[The  difference  of  one  pair  of  arcs  =  difference  of  a  second  pair ;  the 
difference  of  a  third  pair  =  difference  of  a  fourth  pair.  (Explain.) 
Transpose  negative  terms  and  add  correctly,  noting  that  the  sum  of  4 
arcs  =  sum  of  4  others,  and  hence  =  180°.  Half  the  sum  of  these  4  arcs 
measures  the  angle  between  the  bisectors.  (Explain.)  Etc.] 


ff — 


BOOK  II 


111 


52.  If  a  tangent  be  drawn  at  one  end  of  a  chord,  the 
midpoint  of  the  intercepted  arc  will  be  equally  distant 
from  the  chord  and  tangent. 

[Draw  chord  A M  and  prove  the  rt.  &  =  .] 

53.  If  two  circles  are  tangent  at  A  and  a  common 
tangent  touches  them  at  B  and  C,  the  angle  BAG  is  a 
right  angle.     [Draw  tangent  at  A.     Use  219;  251.] 

54.  A   circle  described  on  the  radius  of  another 
circle  as  diameter  bisects  all  chords  of  the  larger  circle 
drawn  from  their  point  of  contact.     To  Prove:    AB  is 
bisected  at  C.     Draw  chord  OC.     (Use  251 ;  213.) 

55.  If  the  opposite  angles  of  a  quadrilateral  are 
supplementary,  a  circle  can  be  drawn  circumscribing  it. 

To  Prove :    A  O  can  be  drawn  through  A,  B,  C,  P. 

Proof:  A  O  can  be  drawn  through  A,  B,  C  (?).     It 
is   required   to    prove  that  it  will  contain  point   P. 
Z  P  +  Z  B  are  supp.  (?).   .',  must  be  meas.  by  half  the 
entire  circumf.     Z  B  is  meas.  by  £  arc  ADC  (?).     Hence 
Z  P  is  meas.  by  \  arc  ABC.     If  Z  P  is  within  or  without 
the  circumf.  it  is  not  meas.  by  \  arc  ABC.     (Why  not?) 

56.  The  circle  described  on  the  side  of  a  square,  or  of 
a  rhombus,  as  a  diameter  passes  through  the   point  of 
intersection  of  the  diagonals.     [Use  141 ;  148.] 

57.  The  line  joining  the   vertex  of  the  right  angle 
of  a  right  triangle  to  the  point  of  intersection  of  the 
diagonals  of  the  square  constructed  upon  the  hypotenuse 
as  a  side,  bisects  the  right  angle  of  the  triangle. 

Proof :    Describe  a  O  upon  the  hypotenuse  as  diameter 
and  use  148 ;  209 ;  249. 

58.  If  two  secants,  PAB  and  PCD,  meet  a  circle  at  A,  B,  and  C, 
D  respectively,  the  triangles  PBC  and  PAD  are  mutually  equiangular. 


59.  If  PAB  is  a  secant  and  PM  is  a  tangent  to 

a  circle  from  P,  the  triangles  PAM  andP  BM  are    3 
mutually  equiangular. 

60.  If  two  circles  intersect  and  a  line  be  drawn 
through  each  point  of  intersection  terminating  in 
the  circumferences,  the  chords  joining  these  extrem- 
ities will  be  parallel.     [Draw  RS.     Z  A  is  supp. 
of  Z  RSC  (?).     Finally  use  104.] 


112 


PLANE   GEOMETRY 


61.  If  two  equal  chords  intersect  within  a  circle, 

(1)  One  pair  of  intercepted  arcs  are  equal. 

(2)  Corresponding  parts  of  the  chords  are  equal. 

(3)  The   lines  joining  their  extremities    (in  order) 
form  an  isosceles  trapezoid. 

(4)  .The  radius  drawn  to  their  intersection  bisects  their  angle 

62.  If  a  secant  intersects  a  circumference  at  D  and  E, 
PC  is  a  parallel  chord,  and  PR  a  tangent  at  P  meeting 
secant  at  R,  the  triangles  PCD  and  PRD  are  mutually 
equiangular.  [Z  R  and  Z  CDP  are  measured  by  ^  arc  PC. 
(Explain.)     Etc.] 

63.  If  a  circle  be  described  upon  one  leg  of  a  right 
triangle  as  diameter  and  a  tangent  be  drawn  at  the 
point  of   its  intersection  with  the   hypotenuse,  this 
tangent  will  bisect  the  other  leg. 

[Draw  OP  and  OD.  CD  is  tangent  (?).  OD 
bisects  arc  PC  (?)  (220).  ZCOD  =  ^A  (?)  (257). 
/.  OD  is  ||  to  AB  (?).  Etc.] 

64.  If  an  equilateral  triangle  ABC  is  inscribed  in  a 
circle  and  P  is  any  point  of  arc  AC,  AP  +  PC  =  BP. 
[Take  PN=PA;   draw  AN.     &ANP  is  equilateral. 
(Explain.)     &ANB  =  A  APC  (?).     Etc.] 

65.  If  two  circles  are  tangent  internally  at  C,  and  a 
chord  AB  of  the  larger  circle  is  tangent  to  the  less 
circle  at  M,  the  line  CM  bisects  the  angle  A  CB. 

[Draw  tangent  CX  and  chord  RS.  Z  RSC  = 
Z  BCX  =  Z  A.  .:  AB  is  ||  to  RS.  (Explain.) 
Then  use  233.  Etc.] 

66.  If  two  circles  intersect  at  A   and   C  and 
lines  be  drawn  from  any  point  P,  in  one  circum- 
ference, through  A  and  C  terminating  in  the  other 
at  points  B  and    D,   chord  BD   will  be   of  constant  p 
length  for  all  positions  of  point  P. 

[Draw  BC.     Prove  Z  BCD,  the  ext.  Z  of 
=  a  constant.     Etc.] 

67.  The  perpendiculars  from  the  vertices  of 
a  triangle  to  the  opposite  sides  are  the  bisectors 
of  the  angles  of  the  triangle  formed  by  joining 
the  feet  of  these  perpendiculars. 

To  Prove :  BS  bisects  Z  RST,  etc. 


BOOK  II  113 

Proof:  If  a  circle  be  described  on  AO  as  diam.,  it  will  pass  through  T 
and  5  (?)  (148).  If  a  circle  be  described  on  OC  as  diam.,  it  will  pass 
through  R  and  S  (?).  /.  /.BAR  =  ^BST  (?)  ;  and  ZBCT  =  ZBSR  (?). 
But  Z  BAR  =  Z  BCT.  (Each  is  the  comp.  of  Z.  ABC.  :.  Etc.) 

68.  If  ABC  is  a  triangle  inscribed  in  a  circle,  BD  is 
the  bisector  of  angle  ABC,  meeting  AC  at  0  and  the 
circumference  at  D,  the  triangles  A  OB  and  C0Z)  are 
mutually  equiangular.  Also  triangles  BOC  and  AOD. 
Also  triangles  BOC  and  ^£Z>.  Also  triangles  A  OD  and 
Also  triangles  BCD  and  COD. 


69.  If  two  circles  intersect  at  A  and  B,  and  from  P,  any  point  on  one 
of  them,  lines  A  P  and  BP  be  drawn  cutting  the  other  circle  again  at  C 
and  D  respectively,  CD  will  be  parallel  to  the  tangent  at  P. 

70.  If  two  circles  intersect  at  A  and  B,  and  through  B  a  line  be  drawn 
meeting  the  circles  at  R  and  S  respectively,  the  angle  RAS  will  be 
constant  for  all  positions  of  the  line  RS. 

[Prove  Z.  R  +  Z  S  is  constant.     .'.  Z.  RAS  is  also  constant.] 

71.  Two  circles  intersect  at  A  and  through  A  any  secant  is  drawn 
meeting  the  circles  again  at  M  and  N.    Prove  that  the  tangents  at  M  and 
N  meet  at  an  angle  which  remains  constant  for  all  positions  of  the 
secant. 

[Prove  the  angle  between  these  tangents  equal  to  the  angle  between 
the  tangents  to  the  circles,  at  A  .] 

72.  Three  unequal  circles  are  each  externally  tangent  to  the  other 
two.    Prove  that  the  three  tangents  drawn  at  the  points  of  contact  of 
these  circles  meet  in  a  common  point. 

73.  Two  equal  circles  intersect  at  A  and  B,  and  through  A  any 
straight  line  MAN  is  drawn,  meeting  the  circumferences  at  M  and  N 
respectively.     Prove  chord  BM  =  chord  BN. 

74.  If  the  midpoint  of  the  arc  subtended  by  any  chord  be  joined  to 
the  extremities  of  any  other  chord, 

(1)  The  triangles  formed  will  be  mutually  equiangular. 

(2)  The  opposite  angles  of  the  quadrilateral  thus  formed  will  be  sup- 
plementary. 

75.  Two  circles  meet  at  A   and  B  and  a  tangent  to  each  circle  is 
drawn  at  A,  meeting  the  circumferences  at  R  and  S  respectively;  prove 
that  the  triangles  ABR  and  ABS  are  mutually  equiangular. 


114 


PLANE   GEOMETRY 


76.  What  is  the  locus  of  points  at  a  given  distance  from  a  given 
point?     Prove.     (Review  179  and  180  now.) 

77.  What  is  the  locus  of  the  midpoints  of  all  the  radii  of  a  given 
circle  ?    Prove. 

78.  What  is  the  locus  of  the  midpoints  of  a  series  of  parallel  chords 
in  a  circle  V     Prove. 

79.  What  is  the  locus  of  the  midpoints  of  all  chords  of  the  same 
length  in  a  given  circle  ?    Prove. 

80.  What  is  the  locus  of  all  points  from  which  two  equal  tangents 
can  be  drawn  to  two  circles  which  are  tangent  to  each  other? 

81.  What  is  the  locus  of  all  the  points  at  a  given  distance  from  a  given 
circumference?    Discuss  if  the  distance  is  >  radius.     If  it  is  less. 

82.  What  is  the  locus  of  the  vertices  of  the  right  angles  of  all  the 
right  triangles  that  can  be  constructed  on  a  given  hypotenuse  ?    Prove. 

83.  What  is  the  locus  of  the  vertices  of  all  the  triangles  which  have  a 
given  acute  angle  (at  that  vertex)  and  have  a  given  base  ?     Prove. 

84.  A  line  of  given  length  moves  so  that  its  ends  are  in  two  perpen- 
dicular lines  ;  what  is  the  locus  of  its  midpoint  ?     Prove. 

[Suppose  AB  represents  one  of  the  positions  of 
the  moving  line.  Draw  OP  to  its  midpoint.  In  all 
the  positions  of  AB,  OP  =  1  AB  =  a  constant  (148). 

.'.  P  is  always  a  fixed  distance  from  0.     Etc.] 

85.  What  is  the  locus  of  the  midpoints  of  all  the 

chords  that  can  be  drawn  through  a  fixed  point  on  a  given  circumfer- 
ence ?    Prove.  A 

[Suppose  AB  represents  one  of  the  chords  from  B 
in  circle  0,  with  radius  OB ;  and  P  the  midpoint  of 
AB.  Draw  OP.  Z  P  is  a  rt.Z  (?).  That  is,  where  - 
ever  the  chord  may  be  drawn,  /.  P  is  a  rt.  Z. 

/.  locus  of  P  is,  etc.] 

86.  A  definite  line  which  is  always  parallel  to  a  given  line  moves  so 
that  one  of  its  extremities  is  on  a  given  circumference ;  find  the  locus  of 
the  other  extremity. 

[Suppose  CP  represents  one  position  of  /"  xP 
the  moving  line  CP.  Draw  OQ  =  and  II 
to  CP  from  center  0.  Join  OC  and  PQ. 
Wherever  CP  is,  this  figure  is  a  O  (?).  Its 
sides  are  of  constant  length  (?).  That  is,  P 
is  always  a  fixed  distance  from  Q,  etc.]  A B 


BOOK  ii  115 

CONSTRUCTIONS 

258.  Heretofore  the  figures  we  have  used  have  been  assumed.     We 
have  supposed  such  auxiliary  lines  to  be  drawn  as  conditions  required. 
No  methods  have  been  given  for  drawing  any  lines,  and  only  our  three 
postulates  have  been  assumed  regarding  such  construction.    But  the  lines 
that  have  been  used  were  drawn  as  aids  toward  establishing  truths,  and 
precise  drawings  have  not  been  essential.     The  following  simple  methods 
for  constructing  lines  are  given  that  mathematical  precision  may  be  em- 
ployed if  necessary  in  drawing  diagrams  of  a  more  complex  nature.    The 
pupil  should  be  very  familiar  with  the  use  of  the  ruler  and  compasses. 

259.  A   geometrical   construction   is   a   diagram  made  of 
points  and  lines. 

260.  A  geometrical  problem  is  the  statement  of  a  required 
construction.     Thus:  "it  is  required  to  bisect  a  line"  is  a 
problem.     A  problem  is  sometimes  defined  as  "a  question  to 
be  solved  "  and  includes  other  varieties  besides  those  involved 
in  geometry. 

261.  The  word  proposition  is  used  to  include  both  theorem 
and  problem. 

262.  The  complete  solution  of  a  problem  consists  of  jive 
parts  : 

I.    The  Given  data  are  to  be  described. 
II.    The  Required  construction  is  to  be  stated. 

III.  The  Construction  is  to  be  outlined. 

This  part  usually  contains  the  verb  only  in  the  imperative. 
No  reasons  are  necessary  because  no  statements  are  made. 
The  only  limitation  in  this  part  of  the  process  is,  that  every 
construction  demanded  shall  have  been  shown  possible  by 
previous  constructions  or  postulates.  (See  32;  33;  199.) 

IV.  The  Statement  that  the  required   construction   has 
been  completed. 

V.    The  Proof  of  this  declaration. 

Sometimes  a  discussion  of  ambiguous  or  impossible  in- 
stances will  be  necessary. 


116 


PLANE   GEOMETRY 


263.  NOTES.   (1)  A  straight  line   is  determined  by  two 
points. 

(2)  A  circle  is  determined  by  three  points. 

(3)  A  circle  is  determined  by  its  center  and  its  radius. 
Whenever  a  circumference,  or  even  an  arc,  is  to  be  drawn,  it 
is  essential  that  the  center  and  the  radius  be  mentioned. 

(4)  "Q.  E.F."=  Quod   erat  faciendum  =  "  which  was  to  be 
done."     These  letters  are  annexed  to  the  statement  that  the 
construction  which  was  required  has  been  accomplished. 

264.  PROBLEM.    It  is  required  to  bisect  a  given  line. 
Given  :  The  definite  line  AB. 

Required  :  To  bisect  AB. 

Construction  :  Using  A  and  B 
as  centers  and  one  radius,  suffi- 
ciently long  to  make  the  cir-  __ 
cumferences  intersect,  describe 
two  arcs  meeting  at  E  and  T. 
Draw  ET  meeting  AB  at  M.  \  I  / 

Statement:  Point  M  bisects  /£T 

AB.  Q.E.F. 

Proof :  E  is  equally  distant  from  A  and  B  (?)  (201). 

T  is  equally  distant  from  A  and  B  (?). 

Hence,  ET  is  the  J_  bisector  of  AB  (?)  (70). 

That  is,  M  bisects  AB. 


\ 


Q.E.D. 


265.   PROBLEM.    To  bisect  a  given  arc. 
Given  :  Arc  AB  whose  center 
is  o. 

Required  :  To  bisect  arc  AB. 

Construction :     Draw   chord 
AB ;   using  A  and  B  as  centers  C 
and  any  sufficient  radius,  de- 
scribe arcs  meeting  at  C.   Draw 
0(7  cutting  arc  AB  at  M. 


BOOK  II 


117 


Statement :  The  point  M  bisects  arc  AB.  Q.E.F. 

Proof :   O  and  c  are  each  equally  distant  from  A  and  B  (201). 
.-.  OC  is  the  J_  bisector  of  chord  AB  (?)  (70). 
/.  M  bisects  arc  AB  (?)  (213).  Q.E.D. 


266.   PROBLEM.    To  bisect  a  given  angle. 

Given:    Z  LON. 

Required :    To  bisect  Z  LON. 

Construction :  Using  O  as  a 
center  and  any  radius,  draw  arc  ( 
AB,  cutting  LO  at  A  and  NO  at 
B.  Draw  chord  AB.  Using  A 
and  B  as  centers  and  any  suf- 
ficient radius,  draw  two  arcs 
intersecting  at  S.  Draw  OS  meeting  arc  AB  at  Jkf. 

Statement :    OS  bisects  Z  LON. 


\/s 


Q.E.F. 


Proof  :    O  and  8  are  each  equally  distant  from  A  and  B  (?). 
.*.  OS  is  the  J_  bisector  of  chord  AB  (?). 
.-.  M  bisects  arc  AB  (?)  (213).  .-.  Z  AOM  =  Z  BOM  (?)  (207). 
That  is,  OS  bisects  Z  LON.  Q.E.D. 

267.   PROBLEM.   At  a  fixed  point  in  a  straight  line  to  erect  a  perpen- 
dicular to  that  line. 

Given  :   Line  AB  and  point 
P  within  it. 

Required:   To  erect  a  line 
_L  to  AB  at  P. 

Construction  :    Using  P  as  A— £f 

a   center    and   any    radius, 

draw  arcs  meeting  AB  at  C  and  D.  Using  C  and  D  as  centers 
and  a  radius  longer  than  before,  draw  arcs  meeting  at  S. 
Draw  PS. 

Statement :    PS  is  J_  to  AB  at  P.  Q.E.F. 

Proof :    PS  is  the  _L  bisector  of  CD  (?)  (70).  Q.E.D, 


B 


118 


PLANE   GEOMETRY 


Another  Construction :  Using  any  point  O,  without  AB^  as 
center,  and  OP  as  radius, 
describe  a  circumference, 
cutting  AB  at  P  and  E. 
Draw  diameter  EOS. 
Join  SP. 

Statement :  SP  is  J.  to 
AB  at  p.  Q.E.F. 

Proof:    Segment   SPE  A  EX.  ...-••  p 

is  a  semicircle  (?)  (204). 

/.  Z  SPE  is  a  rt.  Z  (?)  (251).     .-.  SP  is  J.  to  AB  (?)  (17). 

268.   PROBLEM.   Through  a  point  without  a  line  to  draw  a  perpen- 
dicular to  that  line.  P 

Given :    Line     AB    and 
point  P  without  it. 

Required :    (?). 

Construction :    Using    P 
as  a  center  and  any  suffi-     A. 
cient   radius,   describe    an 


.M 


\ 


N.. 


arc  intersecting  AB  at  M  and  N.  Using  M  and  N  as  centers 
and  any  sufficient  radius,  describe  arcs  intersecting  each  other 
at  C.  Draw  PC. 

Statement:   PC  is  _L  to  AB  from  P.  Q.E.F. 

Proof :    PC  is  the  _L  bisector  of  MN  (?)  (70).  Q.E.D. 

269.   PROBLEM.   At  a  given  point  in  a  given  line  to  construct  an  angle 
which  shall  be  equal  to  a  given  angle.  ^  A 

Given  :   Z  A  OB  ;    point  P  in 
line  CD. 

Required :    To  construct  at  P  o- 
an  Z=Z^405. 

Construction:    Using    O  as  a 
center  with  any  radius,  describe 
an  arc  cutting  OA  at  E  and  OB    c 


BOOK  II  119 

at  F.  Draw  chord  EF.  Using  p  as  a  center  and  OE  as  a 
radius,  describe  an  arc  cutting  CD  at  R.  Using  E  as  a  center 
and  chord  EF  as  a  radius,  describe  an  arc  cutting  the  former 
arc  at  X.  Draw  PX  and  chord  RX. 

Statement :    The  Z  XPD  =  Z  AOB.  Q.E.F. 

Proof:    Chord  EF=  chord  EX  (?)  (201). 

.-.  arc  EF=  arc  RX  (?)  (209).  .-.  Z  XPR  =  Z  O  (?)  (207). 
That  is,  Z  XPD  =  Z  ^LOU.  Q.E.D. 

270.  PROBLEM.   To  draw  a  line  through  a  given  point  parallel  to  a 
given  line. 

Given :    Point  P  and  line  AB.     \ 

Required :    To  draw  through     p\ X 

P,  a  line  II  to  AB. 

Construction:  Draw  any  line     

PN  through  P  meeting  AB  at  N.  N 

On  this  line,  at  P,  construct  Z  NPX  =  Z  ^L^P.     (By  269.) 

Statement:   PX  is  II  to  AB.  Q.E.F. 

Proof:    Z.NPX=£ANP   (?)    (Const.). 

/.  PX  is  II  to  AB  (?)  (101).  Q.E.D. 

271.  PROBLEM.   To  divide  a  line  into  any  number  of  equal  parts. 

Given:    Definite  line  AB.     A         o        N        M        L         B 

•••...     /         /         /          /         / 

Required:    To     divide     it       '""£.       /         /        /        / 
into  five  equal  parts.  "'"•£.        /         /         / 

Construction:     Draw  ">,.       /         / 

w»         %^  •  • 

through  A  any  other  line  AX.  ""p...       / 

On  this  take  any  length  AC  '"g... 

as  a  unit,  and  mark  off  on  AX 
five  of  these  units,  AC,  CD,  DE,  EF,  FG.     Draw  GB. 

Through  F,  E,  D,   C,  draw  II  to  GB,  lines  FL,  EM,  DN,  CO. 

Statement :  Then,  AO  =  ON=  NM=  ML  =  LB.  Q.E.F. 

Proof  :    AC  —  CD  =  DE  =  EF  =  FG  (Const.). 

/  ,  AO  =  ON  =  NM  =  ML  =  LB  (?)  (147).  Q.E.D. 


120 


PLANE   GEOMETRY 


272.   PROBLEM.   To  draw  a  tangent  to  a  given  circle  through  a  given 
point: 

I.  If  the  point  is  on  the  circumference. 

II.  If  the  point  is  without  the  circle. 

I.  Given :  O  O ;  P,  a  point 
on  the  circumference. 

Required :  To  draw  a 
tangent  through  P. 

Construction :  Draw  the 
radius  OP.  Draw  line  AB  J_ 
to  OP  at  P  (by  267). 

Statement :  AB  is  tangent 
to  O  O  at  P.  Q.E.F. 


Proof :    AB  is  J_  to  PO  at  P  (Const.). 
.-.  AB  is  a  tangent  (?)  (215)0 


Q.E.D. 


II.  Given  :  O  O ;  P,  a  point 
without  it. 


\ 


Required :  To  draw  a  tan- 
gent through  P. 

Construction:  Draw  PO; 
bisect  it  at  M  (by  264). 

Using  M  as  a  center  and 
PM  as  a  radius,  describe  a 
circumference  intersecting  O  O  at  A  and  B. 

Draw  PA,  PB,  OA,  OB. 

Statement :    PA  and  PB  are  tangents  through  P.        Q.E.F. 
Proof  :  O  M  passes  through  O  (PJf  =  MO  by  const). 

.-.  Z  P40  is  a  rt.  Z  (?)  (251).    /.  PA  is  tangent  (?)  (215). 
Similarly  PB  is  a  tangent.  Q.E.D. 

Ex.  1.    Show  by  two  distinct  methods  how  to  bisect  a  line. 
Ex.  2.    Show  how  to  construct  the  problem  of  270  by  use  of  268. 


BOOK  II 


121 


273.  PROBLEM.  To  circumscribe  a  circle  about  a  given  triangle. 
Given:  (?).     Required:  (?). 

(See  227.) 

Construction  :  Bisect  AB,  EC, 
AC.  Erect  Js  at  these  midpoints, 
meeting  at  O. 

Using  O  as  a  center  and  OA 
as  radius,  draw  a  circle. 

Statement :  This  O  will  pass 
through  vertices  -4,  B,  and  (7. 

Q.E.F. 

Proof:    [Use  85.] 

274.  PROBLEM.  To  inscribe  a  circle  in  a  given  triangle. 

B 


Given:  (?).  Required:  (?).  Construction:  Draw  the 
three  bisectors  of  the  A  of  A  ABC,  meeting  at  O  (by  266). 
Draw  J§  from  O  to  the  three  sides. 

Using  O  as  a  center  and  one  of  these  Js  as  a  radius,  draw  a  O. 

Statement :    This  O  will  be  tangent  to  the  three  sides  of 

&ABC.  Q.E.F. 

Proof  :  The  bisectors  of  these  angles  meet  in  a  point  and 
the  Js  OL,  OM,  ON  are  equal  (?)  (84). 

/.  the  circumference  passes  through  i,  3f,  N  (?)  (192). 
Therefore  the  three  sides  are  tangent  to  the  O  (?)  (215). 
That  is,  the  O  O  is  inscribed  in  A  ABC  (?)  (234).  Q.E.D. 


122 


PLANE   GEOMETRY 


275.   PROBLEM.  To  construct  a  parallelogram  if  two  sides  and  the 


Q.E.F. 


included  angle  are  given.  w 

Given  :  The  sides  a  and  b  and 
their  included  angle,  x. 

Required :     To    construct    a 
O  containing  these  parts. 

Construction :  Take  a  straight 
line  PQ  =  a. 

At    P    construct   Z  P  =  Z  x.     u 

On    PTF,    the   side    of    this  Z,     b 
take  PR  =  b. 

At  R  draw  RT  \\  to  PQ;  and  at  Q  draw  QZ  ||  to  PW. 

Denote  the  intersection  of  these  lines  by  S. 

Statement :    PQSR  is  the  required  parallelogram, 

Proof  :    First,  it  is  a  parallelogram  (?). 

Second,  it  is  the  required  parallelogram.     (Because  it  con- 
tains the  given  parts.)  Q.E.D. 

276.   PROBLEM.  To  construct  a  segment  of  a  circle  upon  a  given 
line,  as  chord,  which  shall  contain  angles  equal  to  a  given  angle. 

Given  :  Line  AB  and 
Zir'. 

Required :  To  con- 
struct a  segment  upon  AB 
whose  inscribed  angles 
shall  =  Z  K1. 

Construction:  Con- 
struct at  A,  Z.BAC=/.K'. 

Bisect  AB  at  M. 

At  M  erect  OM  _L  to  AB. 

At  A  erect  OA  _L  to 
AC,  meeting  OM  at  o. 

Using  o  as  a  center  and  OA  as  radius,  describe  O  O. 

Statement :    The  A  inscribed  in  AKB  =  Z  K1.  Q.E.F, 


BOOK  II  123 

Proof:  The  circumference  passes  through  B  (?)  (67). 
.-.  AB  is  a  chord  (?).  AC  is  tangent  to  the  O  (?)  (215). 
.-.  ZBACis  measured  by  half  the  arc  AB  (?)  (252). 

Any  inscribed  angle  AKB  is  measured  by  half  the  arc  A  B  (?). 

Therefore,  any  angle  AKB  =  Z  BAG  (?)  (257,  2). 

Consequently,  any  inscribed  angle  AKB  =  Z  Ef  (Ax.  1). 

Q.E.D. 

[If  the  pupil  will  draw  chords  AK  and  BK,  he  will  under- 
stand the  proposition.  These  were  purposely  omitted.] 

277.  PROBLEM.  To  construct  the  third  angle  of  a  triangle  if  two 
angles  are  known. 

Given  :    A  A  and  J5,  two  A 
of  a  A.  A- 

Required :  To  construct  the 
third. 

Construction :  At  point  O  in 
a  line  Its  construct  Z  a  =Z  A.    R~"  o' 

At  point  O  in  or  construct  Z  b  =  Z  B. 

Statement :   The  Z  VOB  =  the  third  Z  of  the  A.         Q.B.F. 

Proof  :    Z.a  +  Z&  +  Z  VOE  =  2  rt.  A  (?)  (46). 

Z^l  +  Z£  +  the  unknown  Z  =  2  rt.  A  (?)  (HO). 

.-.  Z  a  +  Z  6  +  Z  VOB  =  Z.A  +  /.B  +  the  unknown  Z  (?). 

But  Ztf  +  Z5 =  Z  A  +  Z  B  (Const,  and  Ax.  2). 

.*.  Z  VOB       =  the  unknown  Z  (Ax.  2). 
That  is,  ZFOS  =  the  third  Z  of  the  A.  Q.E.D. 


Ex.  1.   To  circumscri.be  a  triangle  about  a  given  circle. 

Ex.  2.  To  construct  the  problem  of  276  if  the  given  angle  is  a  right 
angle  ;  if  it  is  an  obtuse  angle. 

Ex.  3.    To  construct  the  problem  of  273  if  the  given  triangle  is  obtuse. 

Ex.4.    Is  the  problem  of  277  ever  impossible?    Explain. 

Ex.  5.  In  the  figure  of  274,  if  Z  A  =  40°  and  Z£  =  94°,  how  many 
degrees  are  there  in  each  of  the  six  acute  angles  at  0  ?  If  A  LMN 
is  constructed,  how  many  degrees  are  there  in  each  of  its  angles  ? 


124  PLANE   GEOMETRY 

278.  PROBLEM.  To  construct  a  triangle  if  the  three  sides  are  known. 
Given :    Sides 

a,  5,  c  of  a  A. 

0 

Required :  To  tf 
construct  the  A. 

Construction  : 
Draw  ES  =  a. 

Using  E  as  a  R  ' 

center  and  6  as  a  radius,  describe  an  arc ;  using  S  as  a  center 
and  c  as  a  radius,  describe  an  arc  intersecting  the  former  arc 
at  T.  Draw  ET  and  ST. 

Statement :    EST  is  the  required  A.  Q.B.F. 

Proof:    EST  is  a  A  (?)  (23). 

EST  is  the  required  A.    (It  contains  a,  5,  c.)  Q.E.D. 

Discussion  :    Is  this  problem  ever  impossible  ?    When  ? 

279.  PROBLEM.  To  construct  a  triangle  if  two  sides  and  the  included 
angle  are  known. 

Given  :  The  sides  a  and  5, 
and  their  included  Z  C  in  a  A. 

Required:  To  construct 
the  A. 

Construction  :  Draw  CB  =a. 
At  C  construct  /.  BCX  =  given 
Z  C.  On  CX  take  CA  —  b.  Join  AB. 

Statement:  (?).     Proof:    (?). 


NOTE.  The  student  has  probably  observed  that  in  constructions  cer- 
tain lines  and  angles  must  precede  others.  In  such  problems  as  266,  267, 
269,  272,  and  276,  the  order  of  the  successive  steps  is  exceedingly  impor- 
tant. Problems  are  not  so  numerous  in  geometry  as  theorems,  but  it  must 
be  apparent  that  problems  are  instructive,  fascinating,  and  profitable. 

Definition.  If  a  circle  is  described,  touching  one  side  of  a 
triangle  and  the  prolongations  of  the  other  sides,  it  is  called 
an  escribed  circle. 


BOOK  II 


125 


280.   PROBLEM.  To  construct  a  triangle  if  a  side  and  the  two  angles 
adjoining  it  are  known. 

a    -•^•--  •-.-- r-r-.-- — 


—  cf 


Given:  (?).    Required:   (?).  a 


Construction :  Draw  BC  =  a.  At  B  construct  Z  CB X  —  Z  B1 ; 
at  C  construct  Z  BCY  =  Z  Cr.  Denote  the  point  of  intersec- 
tion of  BX  and  CY  by  A. 

Statement:    (?).     Proof:    (?).     Discussion:     (?). 

281.  PROBLEM.  To  construct  a  right  triangle  if  the  hypotenuse  and 
a  leg  are  known. 

Given:  Hypotenuses;  leg b.   A 

Required:    (?). 

Construction:  Draw  an  in- 
definite line  CD  and  at  C  erect 
a  J_  =  b.  Using  A  as  a  center 
and  c  as  a  radius,  describe  an 
arc  cutting  CD  at  B.  Draw  AB. 

Statement:  (?).     Proof:  (?).    Discussion:  (?). 

NOTE.  If  it  is  required  to  construct  a  right  triangle,  having  given  the 
hypotenuse  and  another  part,  it  is  often  advantageous  to  describe  a  semi- 
circle upon  the  given  hypotenuse  as  diameter.  Every  triangle  whose  base 
is  this  diameter  and  whose  vertex  is  on  this  semicircumference  is  a  right 
triangle  (251).  Hence  if  the  triangle  constructed  contains  the  other 
given  part,  it  is  the  required  triangle. 

Ex.  1.  To  construct  the  problem  of  281  by  use  of  the  semicircle. 

Ex.  2.  Discuss  the  constructions  of  279,  280,  and  281  fully. 

Ex.  3.  To  construct  a  triangle  and  its  three  escribed  circles. 

Ex.  4.  To  construct  an  isosceles  right  triangle  having  given  the  hypot- 
enuse. 

Ex.  5.  To  construct  an  isosceles  right  triangle  having  given  the  leg. 

Ex.  6.  To  construct  a  quadrilateral  having  given  the  four  sides  and 
an  angle. 


126 


PLANE   GEOMETRY 


282.  PROBLEM.  To  construct  a  triangle  if  an  angle,  a  side  adjoin- 
ing it,  and  the  side  opposite  it  are  known;  that  is,  if  two  sides  and 
an  angle  opposite  one  of  them  are  known. 

The  known  angle  may  be  obtuse,  right,  or  acute.   Consider : 
First,  If  "side  opposite"  >  "side  adjoining." 
Second,  If  "side  opposite "=  "side  adjoining." 
Third,  If  "side  opposite  "<  "side  adjoining." 
Construction  for  all  of  these :    Draw  an  indefinite  line,  EX, 
and  at  one  extremity  construct  an  Z  =Z  E  ;  take  on  the  side 
of  this  Z  a  distance  from  the  vertex  =  "side   adjoining." 
Using  the  end  of  this  side  as  a  center  and  the  "side  opposite  " 
as  a  radius,  describe  an  arc  intersecting  EX.     Draw  radius 
to  the  intersection  just  found. 

If  the  known  angle  is  obtuse  or  right 

Given :  Z  E,  s.a,  and  s.o.  of  a  A.  \ 

Construction :  As  above. 

Discussion:  Case  I.  s.o.  >s.a. 
The  A  is  always  possible. 

Case  II.  8.o.=  s.  a.  The  A  is 
never  possible  (55  and  112). 

Case  III.    s.  o.  <  s.  a. 

The  A  is  never  possible 
(?)  (122). 

If  the  known  angle  is 
acute. 

Case     I.  s.  o.  >  s.  a. 
The  A  is  always  possible. 

Case  II.  s.o.  =  8. a.  An 
isosceles  A. 

Case  III.   s.  o.  <  s.  a. 

(1)  If  s.o.  <  the  J_  from 
A  to  EX,  the  A  is  never  pos- 
sible. 


BOOK  II 


12? 


A 


•X 


(2)  If  8.0.  =  the  _L  from 
A  to  KX,  A  is  a  rt.A  (216). 

(3)  If  s.o.  >  the  J_  from 
to  JBTJr,  there  are  two  A. 
In   this   instance  the  arc 

described,  using  A  as  cen- 
ter and  "  s.o."  as  radius,  in- 
tersects KX  twice,  at  B 
and  Bf. 

Hence,  A  AKB  and  A  A  KB1 
both  contain  the  three  given 
parts. 

CONCERNING  ORIGINAL   CONSTRUCTIONS 
ANALYSIS 

Many  constructions  are  so  simple  that  their  correct  solution  will 
readily  occur  to  the  pupil.  Sometimes,  as  in  the  case  of  complicated  con- 
structions, one  requires  the  ability  to  put  the  given  parts  together,  one 
by  one. 

The  following  outline  may  be  found  helpful  if  employed  intelligently. 
I.  Suppose  the  construction  made, — that  is,  suppose  the  figure  drawn. 
II.  Study  this  figure  in  search  of  truths  by  which  the  order  of  the 
lines  that  have  been  drawn  can  be  determined.     This  is  essential. 

III.  One  or  more  auxiliary  lines  may  be  necessary. 

IV.  Finally,  construct  the  figure  and  prove  it  correct. 

EXERCISE.  Given  the  base  of  a  triangle,  an  adjacent  acute  angle, 
and  the  difference  of  the  other  sides,  to  construct  the  triangle. 

Given  :  Base  AB ;  Z  A' ;  difference  d. 

Required :  To  construct  the  A. 

[Analysis :  Suppose  A  ABC  is  the  required  A. 
It  is  evident  that  if  CD  =  CB,  they  may  be  sides 
of  an  isos.  A  and  AD  =  d.  This  isosceles  A  may 
be  constructed.] 

Construction:  At  A  on  AB  construct 
Z.EAX.  —  /.A'  and  take  on  AX,  AD  =  d. 
Join  DB.  At  M,  midpoint  of  DB,  draw  MY 
JL  to  DB  meeting  AX  at  C.  Draw  CB. 

Statement :  (?).  Proof :  (?).  Discussion :  (?).     A- 


128  PLANE   GEOMETRY 

ORIGINAL    CONSTRUCTIONS 

1.  To  construct  a  right  angle. 

2.  To  construct  an  angle  containing  45°. 

3.  To  construct  the  complement  of  a  given  angle ;  the  supplement. 

4.  To  construct  an  angle  of  60°.     [See  115.] 

5.  To  construct  an  angle  of  30° ;  of  15° ;  of  120°. 

6.  To  construct  an  angle  of  150°;  of  135°;  of  75°  ;  of  165°. 

7.  To  find  the  center  of  a  given  circle.     [See  214.] 

8.  To  construct  a  tangent  to  a  given  circle,  parallel  to  a  given  line. 
[Draw  a  radius  _L  to  the  given  line.] 

9.  To  construct  a  tangent  to  a  given  circle,  perpendicular  to  a  given 
line. 

10.  To  construct  the  other  acute  angle  of  a  right  triangle  if  one  is 
known. 

11.  To  draw  through  a  given  point  without  a  given  line,  another  line 
which  shall  make  a  given  angle  with  the  line. 

[Draw  a  ||  to  the  given  line  through  the  given  point.] 

12.  To  trisect  a  right  angle. 

13.  To  find  a  point  in  one  side  of  a  triangle  equally  distant  from  the 
other  sides.     [Use  266.] 

14.  To  construct  a  chord  of  a  circle  if  its  midpoint  is  known. 
[Draw  a  radius  through  this  point  and  use  267.] 

15.  To  construct  the  shortest  chord  that  can  be  drawn  through   a 
given  point  within  a  circle. 

Proof:  Draw  any  other  chord  through  the  point,  etc. 

16.  To  construct  through  a  given  point  within  a 
circle  two  chords  each  equal  to  a  given  chord. 

17.  To  construct  in  a  given  circle  a  chord  equal  to 
a  second  chord  and  parallel  to  a  third. 

18.  To  construct  through  a  given  point  a  line 
which  shall  make  equal  angles  with  the  sides  of  a 
given  angle.     [Use  266  ;  268.] 

19.  To  construct  from  a  given  point  in  a  given  o« 
circumference  a  chord  which  shall  be  at  a  given 
distance  from  the  center. 

[How  many  can  be  drawn  from  this  point?] 


BOOK  II 


129 


1.   To  construct  an  isosceles  triangle,  having  given : 

20.  The  base  and  one  of  the  equal  sides. 

21.  The  base  and  one  of  the  equal  angles. 

22.  One  of  the  equal  sides  and  the  vertex-angle. 

23.  One  of  the  equal  sides  and  one  of  the  equal  angles. 
24    The  base  and  altitude  upon  it. 

25.  The  base  and  the  radius  of  the  inscribed  circle. 
[Bisect  the  base;  erect  a  JL  =  radius;  describe  O,  etc.] 

26.  The  base  and  the  radius  of  the  circumscribed  circle. 
[First,  describe  O  with  given  radius  and  any  center.] 

27.  The  altitude  and  the  vertex-angle. 

[Draw  an  indefinite  line  and  erect  a  _L  equal  the  given  altitude.     Bisect 
the  given  Z ;  at  the  end  of  the  altitude  construct  /.  —  \  given  Z,  etc.] 

28.  The  base  and  the  vertex-angle. 

[Find  the  supplement   of  given  Z;  bisect  this;  at  each  end  of  base 
construct  an  Z  —  this  half;  etc.] 

29.  The  perimeter  and  the  altitude. 
Given:     Perimeter  =  AB;       alt.  =  h. 

Required:   (?).  Construction:  Bisect  A B, 
erect  at  M  _L  =  h;  draw  AP  and  BP. 
Bisect  these ;  erect  Js  SC  and  RE ;  etc. 


II.  To  construct  a  right  triangle,  having  given : 

30.  The  two  legs. 

31.  One  leg  and  the  adjoining  acute  angle. 

32.  One  leg  and  the  opposite  acute  angle. 

33.  The  hypotenuse  and  an  acute  angle. 

34.  The  hypotenuse  and  the  altitude  upon  it. 

35.  The  median  and  the  altitude  upon  the  hypote- 
nuse. [Same  as  No.  34.] 

36.  The  radius  of   the   circumscribed   circle  and 
a  leg. 

37.  The  radius  of  the  inscribed  circle  and  a  leg. 
Given:  Radius  =  r;    leg  =  CM.     Required:    (?). 

Analysis  :  Consider  that  A  EC  is  the  completed  figure ; 
CNOM  is  a  square,  whose  vertex  O  is  the  center  of 
the  circle,  and  side  ON  is  the  given  radius.  AB  is  tan- 
gent from  A.  Construction  :  On  CA  take  CN=  r  and 
construct  square,  CNOM.  Prolong  CM  indefinitely. 


Describe  O,  etc. 


130  PLANE   GEOMETRY 

38.  One  leg  and  the  altitude  upon  the  hypotenuse. 

39.  An  acute  angle  and  the  sum  of  the  legs. 
Given:  AD  =  sum;   Z  K.    Required:   (?). 

Construction  :  At  A  construct  Z  A  =  Z  K;  at 
D  construct  Z  D  =  45°,  the  sides  of  these  A 
intersecting  at  B.  Draw  EC  ±  to  AD ;  etc. 

40.  The  hypotenuse  and  the  sum  of  the  legs. 
[Use  A   as  center,  hypotenuse  as  radius,  etc.] 

41.  The  radius  of  the  circumscribed  circle  and  an  acute  angle. 

42.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 
Construction:   Take    CS   on  indefinite  line 

ZA  =  r.  On  CS  construct  square  CSOM.  At 
O  construct  Z  MOX  =  /.K.  Draw  radius  OT 
-L  to  OX.  Draw  tangent  at  T.  Proof:  A  ABC 
is  a  rt.  A  and  it  is  the  rt.  A.  (Explain.)  2""~ 


III.  To  construct  an  equilateral  triangle,  having  given : 

43.  One  side. 

44.  The  altitude. 

45.  The  perimeter. 

46.  A  median. 

47.  The  radius  of  the  inscribed  circle. 

[Draw  circle  and  radius;  at  center  construct  Z.ROS 
120°  and  Z  ROT  =  120°;  etc.] 

48.  The  radius  of  the  circumscribed  circle. 


IV.   To  construct  a  triangle,  having  given  : 

49.  The  base,  an  angle  adjoining  it,  the  altitude  upon  it. 

50.  The  midpoints  of  the  three  sides. 
[Draw  RS,  R  T,  ST,  etc.] 

51.  One  side,  altitude  upon  it,  and  the  radius 
of  the  circumscribed  circle. 

Construction :    Draw  O  with  given  radius  and     A  ~~r~~       "^ ! fc 

any  center.     Take  chord  =  given  side  ;  etc. 

52.  One  side,  an  adjoining  angle,  and  the  radius  of  the  circumscribed 
circle. 

53.  Two  sides  and  the  altitude  from  the  same  vertex. 
Construction:    Erect  _L  =  altitude,  upon  an  indefinite  line.     Use  the 

end  of  this  altitude  as  center  and  the  given  sides  as  radii;  etc. 


BOOK  IJ 


131 


54.  One  side,  an  angle  adjoining  it,  and  the 
sum  of  the  other  two  sides. 

Construction:  At  A  construct  Z  BA X  =  given 
ZK.  On  AX  take  AD  —  s^  draw  DB ;  bisect 
DB  at  M,  etc. 

55.  Two  sides  and  the  median  to  the  third  side. 
Given :  a,  b,  m.    Construction :    Construct  A  A  BR 

whose  three  sides,  AB  =  a,  BR  —  &,  A R  =  2  m. 
Draw^C  II  ioBR  and  RC  II  to  AB  meeting  at 
C.  Draw^C.  Statement:  (?).  Proof:  (V). 

56.  A  side,  the  altitude  upon  it,  and  the  angle 
opposite  it. 

Given :  Side  —  AB,  alt.  =  h ;  opposite  Z  =  Z  C". 

Construction:  Upon  AB  construct  segment 
ACB  which  will  contain  A  =  Z  C'  (by  276). 
At  A  erect  J.R  JL  to  ^4£  and  =  A;  etc. 

57.  A  side,  the  median  to  it,  the  angle  oppo- 
site it. 

[Statement:  A  ABC  is  the  required  A.] 

58.  One  side  and  the  altitude  from  its  extremi- 
ties to  the  other  sides.  m 

Given:  Side  =  AB,  altitudes  x  and  y. 

Construction:  Bisect  A B;  describe  a  semicircle.  Using 
A  as  center  and  x  as  radius,  describe  arc  cutting  the 
semicircle  at  R',  etc. 

59.  Two  sides  and  the  altitude  upon  one  of  them.         A         M     ""B 
[Given :  Sides  =  A  B  and  BC ;  alt.  on  EC  =  r.]  x 

60.  One  side,  an  angle  adjoining  it,  and  the  radius  of  the  inscribed 
circle. 

Construction:   Describe   O  with  given  radius,  any  center. 
Construct  central  Z  =  given  Z.     Draw  two  tangents  ||  to  these  radii. 


V.   To  construct  a  square,  having  given : 

61.  One  side. 

62.  The  diagonal. 

63.  The  perimeter. 

64.  The  sum  of  a  diagonal  and  a  side. 


132  PLANE  GEOMETRY 

VI.  To  construct  a  rhombus,  having  given: 

65.  One  side  and  an  angle  adjoining  it. 

66.  One  side  and  the  altitude. 

67.  The  diagonals. 

68.  One  side  and  one  diagonal.     [Use  278.] 

69.  An  angle  and  the  diagonal  to  the  same  vertex. 

70.  An  angle  and  the  diagonal  between  two  other  vertices. 

71.  One  side  and  the  radius  of  the  inscribed  circle. 


VII.  To  construct  a  rectangle,  having  given: 

72.  Two  adjoining  sides. 

73.  A  diagonal  and  a  side. 

74.  One  side  and  the  angle  formed  by  the  diagonals. 

75.  A  diagonal  and  the  sum  of  two  adjoining  sides.     [See  No.  40.] 

76.  A  diagonal  and  the  perimeter. 

77.  The  perimeter  and  the  angle  formed 
by  the  diagonals. 

Construction :  Bisect  the  perimeter  and 
take  AB  =  half  it.    Bisect  Z  K.    MA  con-  K 
struct  Z  BAX  =  half  Z  K.    Etc. 


VIII.   To  construct  a  parallelogram,  having  given : 

78.  One  side  and  the  diagonals.     [Use  137  and  278.] 

79.  The  diagonals  and  the  angle  between  them. 

80.  One  side,  an  angle,  and  the  diagonal  not  to  the  same  vertex. 

81.  One  side,  an  angle,  and  the  diagonal  to  the  same  vertex. 

82.  One  side,  an  angle,  and  the  altitude  upon  that  side. 

83.  Two  adjoining  sides  and  the  altitude. 


IX.  To  construct  an  isosceles  trapezoid,  having  given: 

84.  The  bases  and  an  angle  adjoining  the  larger  base. 

85.  The  bases  and  an  angle  adjoining  the  less  base. 

86.  The  bases  and  the  diagonal. 

87.  The  bases  and  the  altitude. 

88.  The  bases  and  one  of  the  equal  sides. 

89.  One  base,  an  angle  adjoining  it,  and  one  of  the  equal  sides. 

90.  One  base,  the  altitude,  and  one  of  the  equal  sides. 


BOOK  II  133 

91.  One  base,  the  radius  of  the  circumscribed  circle,  and  one  of  the 
equal  sides.     [First,  describe  a  O.] 

92.  One  base,  an  angle  adjoining  it,  and  the  radius  of  the  circum- 
scribed circle. 

93.  The  bases  and  the  radius  of  the  circumscribed  circle. 

94.  One  base  and  the  radius  of  the  inscribed  circle. 
Construction :   Bisect  the  base  and  erect  a  _L  =  radius ;  etc. 


X.   To  construct  a  trapezoid,*  having  given: 

95.  The  bases  and  the  angles  adjoining  one  of  them. 
Construction:   Take  EC  —  longer  base,  and  on  it  take  ED  =  less  base. 

Construct  A  DEC  (by  280). 

96.  The  four  sides. 

97.  A  base,  the  altitude,  and  the  non-parallel  sides. 
Construction :   Construct  a  A  two  sides  of  which  =  the  given  n  on-  II  sides 

of  the  trapezoid,  and  the  alt.  from  same  vertex  =  given  alt.     (See  No.  53.) 

98.  The  bases,  an  angle,  and  the  altitude. 

Construction:   Construct  d  on  ED,  having  given  altitude  and  Z. 

99.  A  base,  the  angles  adjoining  it,  and  the  altitude. 

100.  The  longer  base,  an  angle  adjoining  it,  and  the  non-parallel  sides. 

101.  The  shorter  base,  an  angle  not  adjoining  it,  and  the  non-parallel 
sides. 


XI.   To  construct  the  locus  of  a  point  which  will  be : 

102.  At  a  given  distance  from  a  given  point. 

103.  At  a  given  distance  from  a  given  line. 

104.  At  a  given  distance  from  a  given  circumference : 
({)  If  the  given  radius  is  <  the  given  distance; 
(ti)  If  the  given  radius  is  >  the  given  distance. 

105.  Equally  distant  from  two  given  points. 

106.  Equally  distant  from  two  intersecting  lines. 


*NOTE.     It  is  evident  that   every  trapezoid  may  be  divided  into 
parallelogram   and  a  triangle  by  drawing  one 
line  (as  BD)  II  to  one  of  the  non-||  sides.     Hence 
the  construction  of  a  trapezoid  is  often  merely 
constructing  a  triangle  and  a  parallelogram. 


134  PLANE   GEOMETRY 

XII.  To  find  (by  intersecting  loci)  *  the  point  P,  which  will  be: 

107.  At  two  given  distances  from  two  given  points.f 

108.  Equally  distant  from  three  given  points. 

109.  In  a  given  line  and  equally  distant  from  two  given  points. 

110.  In  a  given  line  and  equally  distant  from  two  given  intersecting 
lines. 

111.  In  a  given  circumference  and  equally  distant  from  two  given 
points,  f 

112.  In  a  given  circumference  and  equally  distant  from  two  inter- 
secting lines.f 

113.  Equally  distant  from  two  given  intersecting  lines  and  equally 
distant  from  two  given  points,  f 

114.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  given  points.f 

115.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  other  intersecting  lines. f 

116.  Equally  distant  from  two  given  points  and  at  a  given  distance 
from  one  of  them.f 

117.  Equally  distant  from  two  given  intersecting  lines  and  at  a  given 
distance  from  one  of  them.f 

118.  At  a  given  distance  from  a  point  and  equally  distant  from  two 
other  points.f 

119.  At  given  distances  from  two  given  intersecting  lines.f 

120.  At  given  distances  from  a  given  line  and  from  a  given  circum- 
ference, f 

121.  At  given  distances  from  a  given  line  and  from  a  given  point.f 

122.  Equally  distant  from  two  parallels  and  equally  distant  from 
two  intersecting  lines.f 

123.  At  a  given  distance  from  a  given  point  and  equally  distant  from 
two  given  parallels.! 


*  It  is  well  to  draw  the  loci  concerned  as  dotted  lines.    (See  No.  124.) 
t  In  the  Discussion,  include  the  answers  to  questions  like  these  : 

(1)  Is  this  ever  impossible  ?  (i.e.  must  there  always  be  such  a  point  ?) 

(2)  Are  there  ever  two  such  points  ?     When  ? 

(3)  Are  there  ever  more  than  two  ?     When  ? 

(4)  Is  there  ever  only  one  ?    When  ?    Etc. 


BOOK  II  135 

124.  At  a  given  distance  from  a  given  point 
and  equally  distant  from  two  given  intersecting 
lines. 

Can  C  be  so  taken  that  there  will  be  no  point  ? 

Can  C  be  so  taken  that  there  will  be  only  one 
point?  Only  two?  Only  three?  More  than  four  ?  V  I 


XIII.   To  find  (by  intersecting  loci)  the  center  of  a  circle  which  will : 

125.  Pass  through  three  given  points.* 

126.  Pass  through  a  given  point  and  touch  a  given  line  at  a  given 
point.* 

127.  Have  a  given  radius  and  be  tangent  to  a  given  line  at  a  given 
point.* 

128.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a 
given  point.* 

129.  Pass  through  a  given  point  and  touch  two  given  parallel  lines.* 

130.  Touch  two  given  parallels,  one  of  them  at  a  given  point.* 

131.  Have  a  given  radius  and  touch  two  given  intersecting  lines.* 

132.  Have  a  given  radius  and  pass  through  two  given  points.* 

133.  Touch  three  given  indefinite  lines,   no  two   of    them    being 
parallel,  f 

134.  Touch  three  given  lines,  only  two  of  them  being  parallel. 


XIV.   To  construct  a  circle  which  will : 

135.  Pass  through  a  given  point  and  touch  a  given  line  at  a  given 
point. 

136.  Touch  two  given  parallel  lines,  one  of  them  at  a  given  point. 

137.  Pass  through  a  given  point  and  touch  two  given  parallels. 

138.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a 
given  point. 

139.  Have  its  center  in  one  line,  touch  another  line,  and  have  a  given 
radius. 


*  Discussion :     Is  this  ever  impossible  ?    Are  there  ever  two  circles  and 
hence  two  centers  ?    Are  there  ever  more  than  two  ?     Etc, 
t  Four  solutions.     One  is  in  274, 


136 


PLANE   GEOMETRY 


140.  Have  a  given  radius  and  touch  two  given  intersecting  lines. 

141.  Have  a  given  radius  and  pass  through  two  given  points. 

142.  Have  a  given  radius  and  touch  a  given  circumference  at  a  given 
point.     [Draw  tangent  to  the  given  O  at  the  given  point.] 

143.  Have  a  given  radius  and  touch  two  given  circumferences. 

144.  Touch  three  indefinite  intersecting  lines.* 

145.  Touch  two  given  intersecting  lines,  one  of  them  at  a  given  point. 

146.  Touch  a  given  line  and  a  given  cir- 
cumference at  a  given  point. 

Given :   Line  AB  ;  O  (7;  point  P. 

Construction :  Draw  radius  CP.  Draw 
tangent  at  P  meeting  AB  at  R.  Bisect 
£  PRB,  meeting  CP  produced  at  0 ;  etc. 

147.  Be  inscribed  in  a  given  sector.  "       •• 
Construction :   Produce  the  radii  to  meet  the  tangent  at  the  midpoint 

of  the  arc.     In  this  A  inscribe  a  O. 

148.  Have  a  given  radius  and  touch  two  given  intersecting  circles. 

149.  Have  a  given  radius,  touch  a  given  line,  and  a  given  circumference. 

150.  Touch  a  given  line  at  a  given  point 
and  touch  a  given  circumference. 

Given:   Line  A  B;  point  P;  O  C. 

Construction :  At  P  erect  PX  ±toAB,  and 
extend  it  below  AB,  so  PR  =  radius  of  O  C. 

Draw  CR  and  bisect  it  at  M. 

Erect  MY  _L  to  CR  at  AT,  meeting  PX  at 
0;  etc. 


B 


151.  What  is  the  locus  of  the  vertices  of  all  right  triangles  having 
the  same  hypotenuse  ? 

152.  Through  a  given  point  on  a  given  circumference  to  draw  two 
equal  chords  perpendicular  to  each  other. 

153.  To  draw  a  line  of  given  length  through  a  given  point  and  ter- 
minating in  two  given  parallels. 

Construction :  Use  any  point  of  one  of  the  Us  as  center  and  the 
given  length  as  radius  to  describe  an  arc  meeting  the  other  ||.  Join 
these  two  points.  Through  the  given  point  draw  a  line  II,  etc. 


*  Four  solutions.     One  is  in  274. 


BOOK  It 


137 


154.  To  draw  a  line,  terminating  in  the 
sides  of  an   angle,  which  shall  be  equal  to 
one  line  and  parallel  to  another. 

Statement :  RS  is  =  a  and  II  to  x. 

155.  To  draw  a  line  through  a  given  point 

within  an  angle,  which  shall  be  terminated  by  the  sides  of  the  angle  and 
bisected  by  the  point. 

Construction:  Through  P  draw  PD 
II  to  AC.  Take  on  AB,  DE  =  AD. 
Draw  EPF;  etc. 

156.  To  circumscribe  a  circle  about  a 
rectangle. 

157.  To  construct  three  circles  having  the  vertices  of  a  given  triangle 
as  centers  so  that  each  touches  the  other  two.  C 

Construction :   Inscribe  a  O  in  the  A ;  etc. 

158.  To  construct  within    a  circle    three    equal 


circles  each  of  which  will  touch  the  given  circle  and 
the  other  two. 

Construction:   Draw  a  radius,  OA,  and  construct 
Z.  A  OB  =  120°  and  Z  A  OC  =  120°.    In  these  sectors  inscribe,  etc. 

159.  Through  a  point  without  a  circle  to  draw 
a  secant  having  a  given  distance  from  the  center. 

160.  To  draw  a  diameter  to  a  circle  at  a  given 
distance  from  a  given  point. 

161.  Through  two  given  points  within  a  circle 
to  draw  two  equal  and  parallel  chords. 

Construction  :  Bisect  the  line  joining  the  given 
points  and  draw  a  diameter,  etc. 

162.  To  draw  a  parallel  to  side  BC  of  tri- 
angle ABC,  meeting  AB  in  X  and  AC  in.   Y, 
such  that  XY=YC. 

163.  Find  the  locus  of  the  points  of  contact 
of  the  tangents  drawn  to  a  series  of  concentric 
circles  from  an  external  point. 

164.  Given :  Line  AB  and  points  C  and  D 
on  the  same  side  of  it ;  find  point  X  in   AB 

such  that  Z  A  XC  =  Z  BXD.  j  / 

Construction:   Draw  CE±to  AB  and  pro-  F 

duce  to  F  so  that  EF  =  CE.    Draw  FD  meeting  AB  in  X. 


Draw  CX. 


138 


PLANE   GEOMETRY 


165.  To  draw  from  one  given  point  to  another 
the  shortest  path  which  shall  have  one  point  in 
common  with  a  given  line. 

Statement :  CX  +  XD  is  <  CR  +  RD. 

166.  To  draw  a  line  parallel  to  side  BC  of  tri- 
angle ABC  meeting  AB  at  X  and  AC  at  Y,  so 

Y  =  BX  +  YC. 


Construction :  Draw  bisectors  of  A  B  and  C,  meeting  at  O,  etc. 

167.  To  draw  in  a  circle,  through  a  given  point  of  an  arc,  a  chord 
which  will  be  bisected  by  the  chord  of  the  arc. 

Construction  :  Draw  radius  OP  meeting  chord  at  C. 
Prolong  PO  to  X  so  CX  =  CP.     Draw  XM  II  to  AB 
meeting  Oat  M.     Draw  PM  cutting  AB  at  D;  etc.      A> 
Is  there  any  other  chord  from  P  bisected  by  AB1 

168.  To  inscribe  in  a  given  circle  a  triangle  whose  angles  are  given. 
Construction :    At    the    center    construct   three  A,   doubles   of   the 

given  A. 

169.  To  circumscribe  about  a  given  circle  a  triangle  whose  angles  are 
given. 

Construction :   Inscribe  A  (like  No.  168)  first,  and  draw  tangents  II  to 
the  sides. 

170.  Three  lines  meet  in  a  point;  it  is  required 
to  draw  a  line  terminating  in  the  outer  two  and 
bisected  by  the  inner  one. 

Construction :  Through  any  point  P,  of  OB, 
draw  Us  to  the  outer  lines.  Draw  diagonal  JR S;  etc. 

171.  To  draw  through  a  given  point,  P,  a  line 
which  will  be  terminated  by  a  given  circumference 
and  a  given  line  and  be  bisected  by  P. 

Construction:  Draw  any  line  DX  meeting  AB 
at  D.  Draw  PE  II  to  AB  meeting  DX  at  E.  Take 
EF=ED-,  etc. 

172.  Through  a  given  point  without  a  circle  to 
draw  a  secant  to  the  circle  which  shall  be  bisected 
by  the  circumference. 

Construction :  Draw  arc  at  T,  using  P  as  center 
and  diam.  of  O  0  as  radius.  Using  T  as  center  and 
same  radius  as  before,  describe  circumference  touch- 
ing O  0  at  C  and  passing  through  P.  Draw  PC 
meeting  O  0  at  M. 


BOOK  II 


139 


173.  To  inscribe  a  square  in  a  given  rhombus. 
[Bisect  the  four  A  formed  by  the  diagonals.] 

174.  To  bisect  the  angle  formed  by  two 
lines  without  producing  them  to  their  point 
of  intersection. 

Construction :  At  P,  any  point  in  RS, 
draw  PA  II  to  XY;  bisect  Z  APS  by  PB. 
At  any  point  in  PB  erect  ML  ±  to  PB, 
meeting  the  given  lines  in  M  and  L.  Bisect 
ML  at  D  and  erect  DC  _L  to  ML,  etc. 

175.  To  construct  a  common  external  tangent  to  two  circles. 
Construction :  Using  0  as  a  center  and  a  B 

radius  =  difference  of  the  given  radii,  con- 
struct (dotted)  circle.  Draw  QA  tangent  to 
this  O  from  Q ;  draw  radius  OA  and  produce 
it  to  meet  given  O  at  B.  Draw  radius  QC  \\ 
to  OB.  Join  EC. 

Statement :  BC  is  tangent  to  both  (D. 

Proof :   AB=CQ  (Const.).     AB  is  ||  to  CQ  (?). 

/.  ABCQisz  a  (V). 

But  Z  OA  Q  is  a  rt.  Z  (?) ;  etc. 

176.  To  construct  a  common  internal 
tangent  to  two  circles. 

Construction :  Using  0  as  a  center  and 
a  radius  =  the  sum  of  the  given  radii, 
construct  (dotted)  circle.  Draw  QA  tan- 
gent to  this  O  from  Q ;  draw  radius  OA 
meeting  given  O  at  B,  etc.,  as  above. 


\ 


BOOK  III 

PROPORTION.    SIMILAR  FIGURES 

283.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another, — both  being  of  the  same  kind. 

284.  A  proportion  is  the  statement  that  two  ratios  are  equal. 

285.  The  extremes  of  a  proportion  are  the  first  and  last 
terms. 

The  means  of  a  proportion  are  the  second  and  third  terms. 

286.  The  antecedents  are  the  first  and  third  terms. 
The  consequents  are  the  second  and  fourth  terms. 

287.  A  mean  proportional  is  the  second  or  third  term  of  a 
proportion  in  which  the  means  are  identical. 

A  third  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  identical. 

A  fourth  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  not  identical. 

288.  A  series  of  equal  ratios  is  the  equality  of  more  than 
two  ratios. 

A  continued  proportion  is  a  series  of  equal  ratios  in  which 
the  consequent  of  any  ratio  is  the  antecedent  of  the  next 
following  ratio. 

289.  EXPLANATORY.     A  ratio  is  written  as  a  fraction  or  as  an  indi- 
cated division;   -,  or  a  -r-  b,  or  a:b.     A  proportion   is  usually  written 

d      x 

r  =  -»  or  a  :  b  =  x  :  y,  and  is  read  :  "  a  is  to  &  as  a;  is  to  y."  In  this  pro- 
o  y 

portion  the  extremes  are  a  and  y ;  the  means  are  b  and  x ;  the  antece- 
dents are  a  and  x ;  the  consequents  are  b  and  y ;  and  y  is  a  fourth 
proportional  to  «,  6,  x.  In  the  proportion  a  :  m  —  m  :  z,  the  mean  pro- 
portional is  m,  and  the  third  proportional  is  z. 

140 


BOOK  m  141 


THEOREMS    AND    DEMONSTRATIONS 

290.  THEOREM.   In  a  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Given  :  —  =  —  or  a  :  b  =  x  :  y.     To  Prove  :  ay  =  bx. 
b      y 

Proof:  —  =  —  (Hyp.).    Multiply  by  the  common  denomi- 
b      y 

nator,  #y  and  obtain,  ay  =  bx  (Ax.  3).  Q.E.D. 

291.  THEOREM.    If  the  product  of  two  quantities  is  equal  to  the 
product  of  two  others,  one  pair  may  be  made  the  extremes  of  a  propor- 
tion and  the  other  pair  the  means. 

Given  :  ay  =  bx. 

To  Prove  :  These  eight  proportions  : 

1.  a  :  b  =  x  :  y,  5.  x  :  y  =  a  :  b, 

2.  a  :  x  =  b  :  y,  6.  x  :  a  =  y  :  5, 

3.  b  :  a  =  y  :  x,  1.  y  :  x  —  b  :  a, 

4.  b  :  y  =  a  :  x,  8.  y  :  b  =  x  :  a. 

Proof:  1.  ay  =  bx   (Hyp.).     Divide  each   member  by  by, 

j    i  ,   .     ay      bx  ,  A       ON  ax  -, 

and  obtain  -£•  =  —  -  (Ax.  3).     /.  —  =  -,  or  a  :  b  =  x  :  y.  Q.E.D. 
by      by  by 

2.  Divide  by  xy\  etc.    3.  &£=«?/  (Hyp.).  Divide  by  ax\  etc. 

NUMERICAL  ILLUSTRATION.  Suppose  in  this  paragraph  a  =  4,  b  =  14, 
x  =  6,  y  =  21  ;  the  truth  of  the  above  proportions  can  be  clearly  seen  by 
writing  these  equivalents.  4  x  21  =  14  x  6  (True). 

1.    4  :  14  =  6  :  21  (True)  ;  2.  4  :  6  =  14  :  21  (True)  ;  etc. 

They  will  all  be  recognized  as  true  proportions. 

292.  THEOREM.    In  any  proportion  the  terms  are  also  in  proportion 
by  alternation  (that  is,  the  first  term  is  to  the  third  as  the  second 
is  to  the  fourth). 

Given  :   a  :  b  =  x  :  y.     To  Prove  :  a  :  x=  b  :  y. 

Proof:  a\l>  —  x\y  (Hyp.).     .*.    ay  =  bx  (290). 

Hence,  a:x=b:y  (291).  Q.E.D. 


142  PLANE   GEOMETRY 

293.  THEOREM.   In  any  proportion  the  terms  are  also  in  proportion 
by  inversion  (that  is,  the  second  term  is  to  the  first  as  the  fourth 
term  is  to  the  third). 

[The  proof  is  similar  to  the  proof  of  292.] 

294.  THEOREM.    In  any  proportion  the  terms  are  also  in  proportion 
by  composition  (that  is,  the  sum  of  the  first  two  terms  is  to  the 
first,  or  second,  as  the  sum  of  the  last  two  terms  is  to  the  third, 
or  fourth). 

Given:  a:b-*:,.      To  Prove:  f  «  +  *'« 

\a  +  b:b 

Proof:  a:b  =  x:y  (Hyp.).    .'.  ay  =  bx  (?)  (290). 

Add  ax  to  each,  and  obtain,   ax  +  ay  =  ax  +  bx   (Ax.  2). 

That  is,  a  (x  -f  y)  —  x  (  a  +  5) . 

Hence,  a  +  b:  a  =  x  +  y  :  x  (?)  (291). 

Similarly,  by  adding  by,  a  -f-  b  :  b  =  x  +  y  :  y.  Q.E.D. 

295.  THEOREM.   In  any  proportion  the  terms  are  also  in  proportion 

by  division  (that  is,  the  difference  between  the  first  two  terms  is  to 
the  first,  or  second,  as  the  difference  between  the  last  two  terms 
is  to  the  third,  or  fourth). 

~.  ,  fa  —  b:a  =  x  —  y:x,  or 

Given  :  a :  b  =  x  :  y.     To  Prove  :  \ 

\a-b:b  =  x-y.y. 

Proof:  a:b  =  x:y   (Hyp.).     .-.  ay  =  bx  (?)   (290). 

Subtracting  each  side  from  ax,  ax—  ay  =  ax  —  bx  (Ax.  2). 

That  is,  a  (x  —  y)  =  x  (a  —  6) . 

Hence,  a-bi  a  =  x- y  :  x  (1)  (291). 

Likewise,  a  —  b:b  —  x  —     :.  Q.E.D. 


NOTE  I.    The  proportions  of  294  and  295  may  be  written  in  many 
different  forms  (292,  293) .     Thus,  (1)  a  ±b:a  =  x±y :  x\ 
(2)  a±b:b  =  x±y:y,  (3)  a  ±  b:x±  y  =  a:  x,  etc. 

NOTE  II.  In  any  proportion  the  sum  of  the  antecedents  is  to  the  sum 
of  the  consequents  as  either  antecedent  is  to  its  consequent.  (Explain.) 
Also,  in  any  proportion  the  difference  of  the  antecedents  is  to  the  differ- 
ence of  the  consequents  as  either  antecedent  is  to  its  consequent. 
(Explain.)  Thus:  a  +  x:b+  y  =  a:b  =  x  :  y. 

Also,  a  —  x:b  —  y  =  a:b  =  x:. 


BOOK  m  143 

296.  THEOREM.  In  any  proportion  the  terms  are  also  in  proportion 
by  composition  and  division  (that  is,  the  sum  of  the  first  two  terms 
is  to  their  difference  as  the  sum  of  the  last  two  terms  is  to  their 
difference). 

Given:   ail  =  x:y.    To  Prove  :  2±| 

a  —  o     x  — 

Proof:  i±l  =  £±l(?)(294). 

Ct  X 

)  (295). 


a  x 

Divide  the  first  by  the  second,  ^       =  x-±&  (?).      Q.E.D. 

a  —  o     x  —  y 

297.  THEOREM.   In  any  proportion,  like  powers  of  the  terms  are 
also  in  proportion,  and  like  roots  of  the  terms  are  in  proportion. 

Given  :    a  :  b  =  x  :  y. 

To  Prove:    an  :  bn  =  xn  :  yn  ;   and  3/a  :  1/b  =  tyx  :  tyy. 

Proof  :   [  Write  the  given  proportion  in  fractional  form,  etc.  ] 

298.  THEOREM.   In  two  or  more  proportions  the  products  of  the  cor- 
responding terms  are  also  in  proportion. 

Given  :    a:b  =  x:y,  and  c  :  d  =  I  :  m,  and  e  :/=  r  :  8. 

To  Prove  :    ace  :  bd/=  xlr  :  yms. 

Proof  :    [Write  in  fractional  form  and  multiply.] 

299.  THEOREM.    A  mean  proportional  is  equal  to  the  square  root  of 
the  product  of  the  extremes. 

Given  :  a  :  x  =  x  :  b.     To  Prove  :  x  =  Voi. 
Proof  :    [Use  290  ;  etc.] 

300.  THEOREM.    If  three  terms  of  one  proportion  are  equal  to  the 
corresponding  three  terms  of  another  proportion,  each  to  each,  the  re- 
maining terms  are  also  equal. 

(  a:  b  =  c:  m,  and  ) 
Given  :  I  [.To  Prove  :  m  =  r. 

(  a  :  b  =  c  :  r.  } 

Proof  :    am  =  be  and  ar  =  be  (?)  (290). 

/.  am  —  ar  (Ax.  1).     Hence,   m  =  r  (Ax.  3).  Q.E.D. 


144  PLANE   GEOMETRY 

301.  THEOREM.  In  a  series  of  equal  ratios,  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents  as  any  antecedent  is  to  its 
consequent. 

r,.  a      c      e     g 

Given:  -~ 


To  Prove:    «+*  +  *+f=g  = 
b  +  d  +/+  hid 

Proof  :    Set  each  given  ratio  =  m  ;  thus, 

a  c  e  a 

-  =  <w;  -  =  w;  -  =  m;f  =  m. 

b  d  f  h 

.*.  a=bm,  c  =  dm,  e=fm,  g=hm  (Ax.  3). 


Hence,  =  (Substitution) 

(Factoring) 


=  m  (Canceling). 

^a^c^e^g  (  *       -,. 

b     d    f     h^  Q-E.D. 

EXERCISES 

1.  If  3:4  =  6:a:,  find  x.  2.   If  8  :  12  =  12  :  ar,  find  x. 

3.  Find  a  fourth  proportional  to  6,  7,  and  15. 

4.  Find  a  third  proportional  to  4  and  10. 

5.  If  11:15  =  a;:  25,  find*.  6.    If  4  :  x  =  x:  25,  find  x. 

7.  Find  a  mean  proportional  between  8  and  18. 

8.  If  7:  x  =  35:  48,  find  x. 

9.  Given,  that  5  :  8  =  15  :  24,  write  seven  other  true  proportions  con- 
taining these  same  four  numbers. 

10.  If  5  x  6  =  2  x  15,  write  eight  proportions  with  these  numbers. 

11.  If  7  :  12  =  21  :  36,  write  the  proportion  resulting  by  alternation  ; 
inversion  ;  composition  ;  division  ;  composition  and  division. 

12.  If  6  :  25  =  18  :  75,  write  the  proportions  required  in  No.  11. 

13.  H  x  +  y:  x  —  y=17:7,  write  the  proportions  that  result  by  virtue 
of  composition  ;  division  ;  composition  and  division. 

14.  Apply  301  to  the  ratios,  |  =  |  =  f  =  ^  =  ^. 


BOOK   III  ,  145 

NOTE.  We  have  seen  that  it  is  possible  to  add  two  lines  and  subtract 
one  line  from  another.  Now  it  is  essential  that  we  clearly  understand 
the  significance  implied  by  indicating  the  multiplication  or  the  division 
of  one  line  by  another. 

What  is  actually  done  is  to  multiply  or  divide  the  numerical  measure 
of  one  line  by  the  numerical  measure  of  another.  Thus  if  one  line  is  8 
inches  long  and  another  is  18  inches  long,  we  say  that  the  ratio  of  the 
first  line  to  the  second  is  T8F  or  f  ,  meaning  that  the  less  line  is  four  ninths 
of  the  larger. 

Also,  in  referring  to  the  product  of  two  lines  we  merely  understand 
that  the  product  of  their  numerical  measures  is  intended. 

If  a  line  is  multiplied  by  itself,  we  obtain  the  square  of  the  numerical 
measure  of  the  line.  The  square  of  the  line  AB  is  written  AB2  or 
(AB)'2  and  the  quantity  that  is  squared  is  the  numerical  value  of  the 
length  of  AB. 

In  the  preceding  paragraphs  of  Book  III,  we  have  been  considering 
numerical  magnitudes.  It  should  be  distinctly  understood  that  in  the 
following  geometrical  propositions  and  demonstrations,  the  foregoing 
interpretation  is  implied  in  multiplication  and  division  involving  lines. 

302.  THEOREM.  A  line  parallel  to  one  side  of  a  triangle  divides  the 
other  sides  into  proportional  segments. 

Given  :    A  ABC  and  line  LM  A 

II  to  BC. 

To  Prove: 

AL  :  LB  =  AM  :  MC. 

Proof:   I.   If    the    parts   AL 
and  LB  are  commensurable. 

There  exists  a  common  unit    _  ___  _ 
of  measure  of  AL  and  LB  (238). 
Suppose  this  is  contained  9  times  in  AL  and  5  times  in  LB. 

Then,  g-  I  (Ax.  3). 

Draw  lines  through  the  several  points  of  division  II  to  BC. 
These  will  divide  AM  into  9  parts  and  MC  into  5  parts. 
All  of  these  14  parts  are  equal  (?)  (147). 


Hence,         =      (Ax.  3).        ,.        =        (Ax.  1). 

MC       5  LB       MC  Q.E.D. 


146  PLANE   GEOMETRY 

II.  If  the  parts  AL  and  LB  are 
incommensurable. 

There  does  not  exist  a  common 
unit  (238).  Divide  AL  into  sev- 
eral equal  parts  (by  271).  Ap- 
ply one  of  these  as  a  unit  of 
measure  to  LB.  There  will  be  a 
remainder,  EB,  left  over  (238). 

Draw  ES   ||   to  BC. 

Now  —  =  —  (The  commensurable  case). 
LR      MS 

Indefinitely  increase   the   number   of    equal  parts  of  AL. 
That  is,  indefinitely  decrease  each  part,  the  unit  or  divisor. 
Hence,  the   remainder,  EB,  will  be  indefinitely   decreased. 
(Because  the  remainder  is  <  the  divisor.) 
That  is,  EB  will  approach  zero  as  a  limit, 
and  8G  will  approach  zero  as  a  limit. 

/.  LE  will  approach  LB  as  a  limit    (240), 
and  MS  will  approach  MC  as  a  limit   (240). 

.-.  —  will  approach  —  as  a  limit  (243), 

LE  LB 

and  —  will  approach  —  as  a  limit    (243). 
M^S  MC 


Consequently,         =         (?)  (242).  Q.E.D. 

LB       MC 

303.  THEOREM.  If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  these  sides  and  their  corresponding  segments  are 
proportional.  A 

Given  :  A  ABC',  LM  \\  to  BC. 

To  Prove: 

I.   AB  :  AC  =  AL  :  AM. 

II.    AB:  AC=  LB  :  MC. 
Proof  : 

AL:LB  =  AM:MC  (?)  (302). 


BOOK  III  147 

.'.     I.    AL  +  LB  :  AL  =  A M  +  MC  :  AM  (?)  (294), 
and    II.   AL  +  LB:LB=AM  +  MC :  MC  (?). 

But  AL  +  LB  =  AB  and  AM  +  MC  =  AC  (Ax.  4). 

Therefore,  I.  AB:AL  =  AC:AM  (Ax.  6). 

Hence,  AB  :  AC  =  AL  :  AM  (?)  (292). 

II.    AB:LB  =  AC:MC  (Ax.  6). 

Hence,  AB:AC  =  LB:MC  (?)  (292).  Q.E.D. 

NOTE.   Each  of  these  proportions  may  be  written  eight  ways   (291). 

And  they  may  be  combined,  thus,  = = •  (Ax.  1). 

A  C      A  M.      IvL  C 

304.  Two  lines  are  divided  proportionally,  if  the  ratio  of 
the  lines  is  equal  to  the  ratios  of  corresponding  segments. 

305.  THEOREM.   If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  it  divides  these  sides  proportionally. 

(Because  the  ratio  of  the  sides  =  the  ratio  of  correspond- 
ing segments  (303).     This  theorem  is  the  same  as  303.) 

306.  THEOREM.    Three  or  more  parallels  intercept  proportional  seg- 
ments on  two  transversals. 

.Given:  (?). 

To    Prove  :    AC :  BD  =  CE :  DF 
=  EG :  FH. 


A.  \. 


Proof:  Draw  from  A,  AT  11  to  / 

E 


BH  intersecting  the  Us,  etc. 


In  A  AES,  / 

G 


—  =  —  =  —(?)  (305). 
AS       AR      ES 


.. 

AB       BS       ST 

But,  AR  =  BD,  B8  =  DF,    8T=  FH  (?)  (130). 

Hence,  AC:  BD=  CE:  DF=EG:  FH  (Ax.  6).  Q.E.D. 


148 


PLANE  GEOMETRY 


307.  THEOREM.    If  a  line  divides  two  sides  of  a  triangle  proportion- 
ally, it  is  parallel  to  the  third  side. 

Given:    A  ABC]    line  DE;    the 
proportion  AB  :  AC  =  AD  :  AE. 

To  Prove  :  DE  is  II  to  BC. 

Proof  :  Through  D  draw  DX  II 
to  BC  meeting  AC  at  X. 

.'.AB  :AC=AD  :  AT  (?)  (305). 
But  AB:  AC  —  AD:  AE  (Hyp.).   B 
(300). 


/.  DX  and  DE  coincide  (?)  (39).     That  is,  DE  is  II  to  BC. 

Q.E.D. 

308.  THEOREM.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  which  are  proportional  to  the  other  two 
sides. 

Given:  A  ABC',    B8  the    bi-   PK. 
sector  of  Z  ABC.  \*  \. 

To  Prove  :   A8  :  8C=AB:  BC. 

Proof  :  Through  A  draw  AP  II 
to  BS,  meeting  JSC,  produced, 
at  P. 

Now,  Z  m  =  Z  x  (?)  (98)  and  Z  w  =  Z  2  (?). 

ButZ  w  =  Zrc  (Hyp.). 

.-.Z.x  =  Z.z  (Ax.  1).     Hence  ^B  =  JBP  (?)  (120). 

In  A  CAP,  BS  is  II  to  AP  (Const.). 

.'.  AS  :  SC  =  BP  :  BC  (?)  (302). 

.'.  AS:  SC=  AB  :  BC  (Ax.  6). 

Ex.     If  AS  =  3,  AB  =  4,  £C  =  9,  find  SC. 

Ex.     If  4  C  =  20,  ^  5  =  9,  BC  =  21,  find  A  S  and  5C. 


Q.E.D. 


309.  The  segments  of  a  line,  made  by  one  of  its  points,  are 
the  lines  between  this  point  and  the  extremities  of  the  line. 

Thus,  in  308,  the  point  S  divides  AC  internally,  into  seg- 
ments SA  and  SC. 


BOOK  III 


149 


But,  if  the  point  8r  be  in  the  prolongation  of  the  given  line, 
the  segments  are  still  8rA  and  s'C,  according  to  the  defini- 
tion, and  the  point  81  di- 
vides AC,  externally,  into  s' 

segments  S!A  and  s'C. 

310.  THEOREM.  The  bisector  of  an  exterior  angle  of  a  triangle  di- 
vides the  opposite  side  (externally)  into  segments  which  are  propor- 
tional to  the  other  two  sides. 

Given :  A  ABC\  BS',  the  bisector  of  exterior  Z  ABD,  meet- 
ing AC  (externally)  at  s1 '.  To  Prove :  A81  :  SrC  =  AB  :  BC. 

Proof :  Through  A  draw  AP  II  to  BSf  meeting  BC  at  P. 

Now,  Z  m  =  Z  x  (?).  Q, 


But  Z  m  =  Z  n  (?). 

/.Zz  =  Zs  (?). 

Hence,  AB  =  BP  (?).  In 
A  CB8r,  AP  is  II  to  B8r  (?). 
.'.  A81 :  S'C  =  BP  :  BC  (?)  (305). 
.'.  A8r  :  SfC=AB:BC  ?. 


Q.E.D. 


Ex.     If  AS'  =  10,  AB  =  7,BC  =  16,  find  S'C  and  AC. 
Ex.    If  4(7  =  14,  AB  =  12,  BC  =  19,  find  AS'  and  S'C. 


311.   A  line  is  divided  harmonically  if  it  is  divided  in- 
ternally and  externally  in  the  same  ratio. 

In  308,  the  line  A  C  is  divided  internally  by  5,  in  the  ratio  AB:BC. 
In  310,  the  line  A  C  is  divided  externally  by  S'  in  the  ratio  AB :  BC. 

THEOREM.    The  bisectors  of  the  interior  and  exterior  angles  of  a  tri- 
angle (at  a  vertex)  divide  the  opposite  side  harmonically. 

Given:  A  ABC',  BS  bisecting  /.ABC',  and  BS'  bisecting  Z  ABD. 
To  Prove:   AS:  SC  =  AS':  S'C.  D' 

Proof:  41=  4|  (?)  (308). 


_ 
SC      S'C 


150 


PLANE   GEOMETRY 


312.  Similar  polygons   are   polygons   that   are   mutually 
equiangular  and  whose  homologous  sides 

are  proportional.  That  is,  every  pair  of 
homologous  angles  are  equal ;  and  the  ratio 
of  one  pair  of  homologous  sides  is  equal  to 
the  ratio  of  every  other  pair  of  homologous 
sides,  a  :  af  =  b  :  b'  =  c :  c'  =  d :  d'  =  etc. 

Triangles  are  similar  if  they  are  mutually  equiangular  and 
their  homologous  sides  are  proportional. 

313.  THEOREM.    Two  triangles  are  similar  if  they  are  mutually 
equiangular. 

Given:  ^ABC.DEF-,  Z.A  =  ZD,  /.B  =  Z#,  Zc  =  Z*\ 

To  Prove :  The  A  are  similar  (that  is,  that  their  sides  are 
proportional). 

D 


Proof :  Place  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  A  ABC  takes  the  position  of -A  DBS. 

Then  Z  DRS  =  Z.E  (Hyp.).     /.  SS  is  II  to  EF  (?)  (102). 

Hence,  DE  :  DR  =  DF  :  DS  (?)  (305). 

That  is,  DE:AB  =  VF:  AC  (Ax.  6). 

Likewise,  by  placing  Z  B  upon  Z  E,  we  may  prove  that, 
DE  :  AB  =  EF  :  BC. 

.'.  DE  :  AB  =  DF  :  AC  =  EF  :  BC  (Ax.  1). 

Therefore,  the  A  are  similar  (?)  (312).  Q.E.D. 


BOOK  III 


151 


314.  THEOREM.    Two  triangles  are  similar  if  two  angles  of  one  are 
equal  to  two  angles  of  the  other.     [See  117  and  313.] 

315.  THEOREM.    Two  right  triangles  are  similar  if  an  acute  angle 
of  one  is  equal  to  an  acute  angle  of  the  other.    [See  314.] 

316.  THEOREM.  If  a  line  parallel  to  one  side  of  a  triangle  intersects 
the  other  sides,  the  triangle  formed  is  similar  to  the  original  triangle. 

Given  :  MN  II  to  BCin  A  ABC. 

To  Prove:  A  AMN  similar 
to  A  ABC. 

Proof:  Z  A  is  common  to 
both;  /.AMN  =  Z.B;  Z_  ANN 
=  Z  C  (?)  (98). 

.-.  A  are  similar  (?)  (313). 

Q.E.D. 


317.  THEOREM.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other  and  the  sides  including  these  angles  proportional,  the 
triangles  are  similar. 

D 


B  C         E  F 

Given:  A  ABC  and  DEF\  Z^4  =  ZD;  DE:  AB  =  DF:  AC. 
To  Prove:   The  A  similar. 

Proof :  Superpose  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  A  ABC  takes  the  position  of  A  DBS. 
Then  DE:DR  =  DF-.DS  (Hyp.).    .-.  RS  is  II  to  EF(?)  (307). 
/.  A  DRS  is  similar  to  A  DEF  (?)  (316).  Q.E.D. 


152  PLANE   GEOMETRY 

318.   THEOREM.  If  two  triangles  have  their  homologous  sides  pro- 
portional, they  are  similar. 

O 


K 


B  C          E  F 

Given  :   A  ABC  and  DEF,  and  DE :  AB  =  DF:  AC  =  EF  :  BC. 
To  Prove :  A  ABC  similar  to  A  DEF. 

Proof :   On  DE  take  DK  =  AB  ;  and  on  DF  take  DL  =  AC. 
Draw  KL. 

Now,    DE  :  AB  =  DF:  AC  (Hyp.). 

/.  DE  :  DK  =  DF  :  DL  (Ax.  6).     /.  KL  is  II  to  EF  (?)  (307). 

Therefore,  A  DKL  is  similar  to  A  DEF  (?)  (316). 
[Now  A  ABC  is  to  be  proven  equal  to  A  DKL.~\ 

DE :  DK  =  EF :  KL.    (Definition  of  similar  triangles,  312.) 

That  is,  DE :  AB  =  EF  :  KL  (Ax.  6). 

But,  DE:  AB  —  EF:  BC  (Hyp.).      .'.  BC  =  KL  (300). 

Hence,  A  ABC  =  A  DKL  (?)  (58). 

But  A  DKL  has  been  proved  similar  to  A  DEF. 

Therefore,  A  ABC  is  similar  to  A  DEF  (Ax.  6).         Q.E.D. 


Ex.  1.   Are  all  equilateral  triangles  similar  ?     Why  ? 

Ex.  2.   Are  all  squares  similar  ?    Why  ? 

Ex.  3.   Are  all  rectangles  similar?     Why? 

Ex.  4.  The  sides  of  a  triangle  are  7,  8,  and  12,  and  the  longest  side 
in  a  similar  triangle  is  30.  Find  the  other  sides. 

Ex.  5.  In  the  figure  of  311,  if  AB  =  10,  AC  =  14,  BC  =  18,  find  the 
four  segments  of  AC  made  by  S  and  Sr. 

Ex,  6.   Prove  the  theorems  of  142  and  143  by  proportion. 


BOOK  III  153 

319    THEOREM.     If   two  triangles  have   their   homologous    sides 
parallel,  they  are  similar. 

A  9 


E 

Given:  A  ABC  and  DEF;  AB  II  to  DE-,  AC  II  to  HF;  and 
BC  II  to  EF. 

To  Prove  :  A  ABC  similar  to  A  DEF. 

Proof  :  Produce  BC  of  A  ABC  until  it  intersects  two  sides 
of  A  DEF  at  B  and  8. 

Now  Z  B  =  Z  DBS,  and  Z  D£S  =  Z  #(?)  (98). 


Likewise,  Z^OB  =  Z  D£R,  and  Z.DSE  =  Z.  F  (?). 


Therefore,  A  ABC  is  similar  to  A  DEF  (?)  (314).       Q.E.D. 

320.   THEOREM.   If  two  triangles  are  similar  to  the  same  triangle, 
they  are  similar  to  each  other. 

Proof  :  The  three  angles  of  each  of  the  first  two  triangles 
are  respectively  equal  to  the  three  angles  of  the  third  (312). 
Hence,  the  first  two  A  are  mutually  equiangular  (Ax.  1). 
Therefore  they  are  similar  (?)  (313).  Q.E.D. 


Ex.  1.  Let  the  pupil  prove  the  theorem  of  319  if  one  triangle  entirely 
surrounds  the  other. 

Ex.  2.   If  one  side  of  one  triangle  intersects  two  sides  of  the  other. 

Ex.  3.  If  they  are  so  placed  that  no  side  of  either,  when  prolonged, 
intersects  any  side  of  the  other  without  being  prolonged. 

[Prolong  any  side  of  one  and  the  sides  not  ||  to  it  in  the  other.] 


154 


PLANE   GEOMETRY 


321.   THEOREM.    If  two  triangles  have  their  homologous  sides  per- 
pendicular, they  are  similar. 


C 
Given:   A  ABC  and  DEF-,  AB  _L  to  DE-,    AC  J_   to  DF; 

BC  JL  to  EF. 

To  Prove  :  A  ABC  similar  to  A  DEF. 

Proof  :  Through  P,  any  point  in  EF,  construct  PR  II  to  AC, 
meeting  DF  at  M.  At  E,  any  point  in  PM,  draw  ES  II  to  AB, 
meeting  ED  at  N.  Draw  PS  II  to  BC,  meeting  NR  at  S,  forming 
the  A  PRS.  PM  is  JL  to  DF  and  RN  is  _L  to  DE  (?)  .  In  quadrilat- 
eral DMRN,  /.  D  +  Z.  M  +  Z.  MEN  -f  Z  N  =  4  rt.  ^  (?)  (165). 

But,  Z  Jf  _  +  Z  jr=2  rt-.  ^S  (Const.). 

(Ax.  2). 


.'.  Z  J)  +  Z  J/EJVr  =  2  rt.  ^ 

But,  Z  a  +  ^  MEN  =2  rt.  Zs  (?)  (46). 

/.  ZD=Z  a  (?)  (49). 

Similarly,  by  quadrilateral  EPSN,  it  may  be  proved  that 
^=Z  b. 

.'.A  1XEF  is  similar  to  A  PRS  (?)  (314). 
But  A  ABC  is  similar  to  A  PES  (?)   (319). 
.-.A  ABC  is  similar  to  A  DEF  (?)  (320).  Q.E.D. 


Ex.  1.   In  the  figure  of  321,  prove  that  Z  F  =  Z.  c  by  using  48. 
Ex.  2.   Draw  the  figure  for  the  theorem  of  321  if  P  is  taken  on  EF 
prolonged.     Prove  the  theorem  with  this  figure. 

Ex.  3.    Prove  the  same  theorem  if  P  is  taken  at  a  vertex. 
Ex.  4.    Prove,  if  P  is  taken  within  the  triangle  DEF. 


BOOK  III 


155 


322.   THEOREM.   Two  homologous  altitudes  of  two  similar  triangles 
are  proportional  to  any  two  homologous  sides. 

Given :  (?). 

BL         AB          AC         BC 


To  Prove: 


B'L'     A'B'     A'C'      B'C' 


A  L          C       JF  U  C 

Proof:  A  ABC  is  similar  to  A  A'B'C'  (?). 

(312).   .-.  A  ABL  is  similar  to  A  A'B'L'  (315). 

But,  AB        AC        BC 


Hence, 


B 


BL         AB         AC         BC    f  .        -,^ 

^7""7^~7v7~^v^  ^        }' 


A'B' 

BC 

B'C1 


A'C'       B'C' 


(312) 


Q.E.D. 


323.   It  is  evident  that :    In  similar  figures, 

1.  Homologous  angles  are  equal. 

2.  Homologous  sides   are   opposite   equal  angles    (in    tri- 
angles). 

Thus,  shortest  sides  are  homologous.    [Opp.  smallest  A.~\ 
Medium  sidea  are  homologous.     [Opposite  medium   A] 
Longest  sides  are  homologous.     [Opposite  largest  A] 

3.  Homologous  sides  are  proportional. 

The  antecedents  of  this  proportion  belong  to  one  of  the 
similar  figures  and  the  consequents  to  the  other. 


Ex.  1.   Prove  the  theorem  of  322  by  use  of  triangles  BLC  and  B'L'C'. 
Ex.  2.    State  all  the  instances  under  which  two  triangles  are  similar. 
Ex.  3.   In  the  figure  of  322,  if  AB  =  13,  AC  =  15,  BL  =  9,  A'C'  =  20, 
find  A'B'  and  B'L'. 


156 


PLANE   GEOMETRY 


324.  THEOREM.  If  two  parallel  lines  are  cut  by  three  or  more  trans- 
versals which  meet  at  a  point,  the  corresponding  segments  of  the  par- 
allels are  proportional. 

Given:     (?). 

~     n  AE       EF       FB 

To  Prove  :      -  =  —  =  -  -  . 

CG        GH       HD 

Proof  :  In  A  COG,  AE  is  ||  to 
CG  (Hyp.).  /.  A  OAE  is  simi- 
lar to  A  OCG  (?)  (316). 

Likewise,  A  OEF  is  similar 
to  A  OGH  and  A  OFB  is  simi- 
lar to  A  OHD  (316). 


CG       OG 


(323,    3);  also  ^  =  2?  (?). 
GH       OG  ^  - 


...  ££  =  £!  (?).     Likewise,    £^=[^ 

CG       GH  GH      \OH 


HD 


Therefore,         =        =         (?). 
CG        GH       HD 


Q.E.D. 


325.   THEOREM.    If  three  or  more  non-parallel  transversals  intercept 
proportional  segments  on  two  parallels,  they  meet  at  a  point. 


<l 


\ 


\C 


Given:  Transversals  AB,  CD, 
EF;  parallels  AE  and  BF]  pro- 
portion, AC  :  BD  =  CE  :  DF. 

To  Prove  :  AB,  CD,  EF  meet 
at  a  point. 

Proof  :  Produce  BA  and  CD 
until  they  meet,  at  o. 

Draw  OF  cutting  AE  at  X.     B  D 

Now,  AC  :  BD  =  CX  :  DF  (?)  (324). 

But  AC  :  BD  =  CE  :  DF  (Hyp.).      .-.  CX  =  CE  (?)  (300). 

Therefore,  FE  and  FX  coincide  (?)  (39). 

That  is,  FE  produced  passes  through  o.  Q.E.D. 


BOOK  III  157 

326.   THEOREM.   The  perimeters  of  two  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 

D' 


A  B  A1  B' 

Given :  Polygon  R  whose  perimeter  =  P  and  similar 
polygon  S  whose  perimeter  =  Pr. 

To  Prove  :    P  :  Pr  =  AB  :  A' B'  =  BC  :  BfCf  =  etc. 

Proof:    AB  :  ArBf  =  BC  :  BrCf  =  CD  :  C'D'  =  etc.  (323,  3). 

/.  AB  +  BC  +  CD  +  etc.  :  ArBr  +  BrCf  +  Cr Df  +  etc.  = 
AB  :  A'B1  =  BC  :  BfCr  =  etc.  (?)  (301). 

.'.  P  :  P'  =  AB  :  AfBf  =  BC  :  BfCr  =  etc.  (Ax.  6).      Q.E.D. 

Ex.  1.  In  the  figure  of  324,  if  AE  =  EF=FB,  prove  CG  =  GH  = 
HD.  State  this  truth  in  a  theorem. 

Ex.  2.  The  median  of  a  triangle  bisects  every  line  that  is  parallel  to 
the  side  to  which  the  median  is  drawn  and  has  its  extremities  in  the 
other  sides  of  the  triangle. 

Ex.  3.  In  the  figure  of  325,  prove  that  the  line  bisecting  A  C  and  BD 
will  pass  through  point  0. 

Ex.  4.   Prove  the  theorem  of  324  if  the  point  0  is  between  the  parallels. 

Ex.  5.  If  AB  and  CD  are  any  two  parallel  lines  whose  midpoints  are 
R  and  S  respectively,  prove  that  the  lines  AD,  BC,  RS  meet  in  a  point. 

Ex.  6.  Two  homologous  sides  of  two  similar  polygons  are  8  and  15. 
The  perimeter  of  the  less  polygon  is  60.  What  is  the  perimeter  of  the 
larger? 

Ex.  7.  The  perimeters  of  two  similar  polygons  are  30  and  125  re- 
spectively. If  the  shortest  side  of  the  larger  is  8$,  find  the  shortest  side 
of  the  less. 

Ex.  8.  The  sides  of  a  polygon  are  5,  6,  7,  8,  10  respectively.  Find  the 
perimeter  of  a  similar  polygon  whose  medium  side  is  17£. 


158 


PLANE   GEOMETRY 


327.  THEOREM.  If  two  polygons  are  similar,  they  may  be  decom- 
posed into  the  same  number  of  triangles  similar  each  to  each  and 
similarly  placed. 


Given  :    Similar  polygons  BE  and  BfEr. 

To  Prove  :  A  ABC  similar  to  AA'B'C'; 
A  ACD  similar  to  A  A'C'D'I 
A  A  ED  similar  to  A  A'E'D'. 

Proof :    First.    AB  :  AfBf  =  BC  :  BfCf  (323,  3). 

Also,  Z  J?=ZJ?'(323,  1). 

Therefore,  A  ABC  is  similar  to  AA'B'C'  (317). 

Second.     In  A  ABC  and  A'B'C',  -^  =  -^7  (?)  (323,  3). 

In  the  similar  polygons,  -2£  =  -5L  (?)  (323,  3). 

B  C        CD 


Consequently, 


AC 


(Ax.  1). 


A'c'     C'D' 

In  the  polygons,  Z  BCD  =  Z.B'C'D'\ 

In  the  A  ABC  and  ^'s'c',  Z  BCA  =  Z  J?W)    ' 
Hence,  by  subtraction,        Z^lCD  =  /.A' C'D'    (Ax.  2). 
Therefore,  A  ^4 CD  is  similar  to  A  A' C'D'  (?)  (317). 
Third.     AAED    is    proved    similar   to    A  A'E'D'    in    like 
manner.  -Q.E.D. 


BOOK  III 


159 


328.   THEOREM.   If  two  polygons  are  composed  of  triangles  similar 
each  to  each  and  similarly  placed,  the  polygons  are  similar. 


H 


K 


Given  :  A  GUI  similar  to  A  G'H'I'  ; 
A  GIJ  similar  to  A  G'l'j1  ; 
A  GJK  similar  to  A  GrJfEr. 

To  Prove  :    The  polygons  HK  and  H!Kr  similar. 

Proof  :    First.     In  &HGI  and  H'G'I',  /.  H  =  Z  H'  (323,  1). 

Also  in  these  A  Z  HIG  =  /.  H'I'G'    (?)  (323,  1). 

In  A  GIJ  and  G'I'J',    Z  GIJ  =  Z  G'/ J'    (?). 

Adding,  /,  HIJ  = /.  H' l' j'     (Ax.    2). 

Likewise,  ZlJK=  Z.I'J'K';  etc. 

That  is,  the  polygons  are  mutually  equiangular. 

Second.    In   A  GHI  and  G'H'I',  -^L  =  JLL  =  _?L  (323,  3). 

G  H        HI        G I 

In  A  GIJ  and  G'I'J',  -^  =  -4^-.    (?). 


Hence, 


GH 


Gl       IJ 

=  J?I  ^^L  (Ax.  1). 
H'I'     i'j' 


T       J-l,  J^"  «^^  ^"G 

In  the  same  way,  we  may  prove    —r-r  =  — — -.  =  — p- .. 
J  J  ^  I'j'       j'^'      ^'^' 


KG 


GH 


I^J^.^etc.CAx.l). 


That  is,  the  homologous  sides  are  proportional. 
Therefore,  the  polygons  are  similar  (?)  (312). 


Q  E.D. 


160  PLANE   GEOMETRY 

329.  THEOREM.  If  through  a  fixed  point  within  a  circle  two  chords 
be  drawn,  the  product  of  the  segments  of  one  will  equal  the  product  of 
the  segments  of  the  other. 

Given :  Point  O  in  circle  C ; 
chords  AB  and  ES  intersecting 
at  O.  (Review  the  note,  p.  145.) 

To  Prove:    Ao  •  OB  =  BO  -  os. 

Proof  :    Draw  AS  arid  EB. 
In    &AOS    and     EOB,     Z  8  = 
ZJ5  (?)   (250). 


.*.  these  A  are  similar  (?)  (314). 

Hence,    AO  :  EO  =  OS  :  OB    (?)   (323,  3). 

.'.  AO  •  OB  =  EO  -  OS   (?)    (290).  Q.B.D. 

330.  THEOREM.   The  product  of  the  segments  of  any  chord  drawn 
through  a  fixed  point  within  a  circle  is  constant  for  all  chords  through 
this  point.     (See  329.) 

331.  Direct  proportion  and  reciprocal  (or  inverse)  proportion. 

Illustrations.  I.  If  a  man  earns  $4|  each  day,  in  8  days  he  will  earn 
$36.  In  12  days  he  will  earn  $51.  Hence,  8  da.  :  12  da.  =  $36  :  $54  is 
a  proportion  in  which  the  antecedents  belong  to  the  same  condition  or 
circumstance,  and  the  consequents  belong  to  some  other  condition  or 
circumstance.  This  is  called  a,  direct  proportion. 

II.  If  one  man  can  build  a  certain  wall  in  120  days,  8  men  can  build 
it  in  15  days;  or  12  men  in  10  days.  Hence,  8  men  :  12  men  =  10  da.  : 
15  da.  is  a  proportion  in  which  the  means  belong  to  the  same  condition 
or  circumstance,  and  the  extremes  belong  to  some  other  condition  or  cir- 
cumstance. This  is  called  a  reciprocal  (or  inverse)  proportion. 

Definitions.  A  direct  proportion  is  a  proportion  in  which 
the  antecedents  belong  to  the  same  circumstance  or  figure, 
and  the  consequents  belong  to  some  other  circumstance  or 
figure.  Thus  the  ordinary  proportions  derived  from  similar 
figures  are  direct  proportions.  (See  323,  3.) 

A   reciprocal  (or  inverse)   proportion   is   a  proportion   in 


BOOK   III 


161 


which  the  means  belong  to  the  same  circumstance  or  figure, 
and  the  extremes  belong  to  some  other 
circumstance  or  figure. 

Thus,  in  the  adjoining  figure,  a  -  b  = 
x  -  y  (329).  /.  a  :  x  =  y  :  b  (291). 
This  is  a  reciprocal  proportion  because 
the  means  are  parts  of  one  chord,  and  the 
extremes  are  parts  of  the  other  chord. 

332.  THEOREM.   If  through  a  fixed  point  within  a  circle  two  chords 
be  drawn,  their  four  segments  will  be  reciprocally  (or  inversely)  pro- 
portional. 

Proof :  [Identical  with  proof  of  329 ;  omitting  the  last  step.] 

333.  THEOREM.    If  from  a  fixed  point  without  a  circle  a  secant  and  a 
tangent  be  drawn,  the  product  of  the  whole  secant  and  the  external  seg- 
ment will  equal  the  square  of  the  tangent. 

Given:  O  C;  secant 
PAB ;  tangent  PT. 

To  Prove : 

PB  •  PA  =  PT2. 

Proof :  Draw  A  Tand  B  T. 

In  A  PAT  and  PBT 
ZP  =  Z  P  (Iden.). 

Z  PTA  is  measured  by 
1  arc  AT  (?).  Z  B  is 
measured  by  J  arc  AT  (?). 

Therefore,  A  PAT  is  similar  to  A  PBT  (?). 

Hence,   PB  :  PT  =  PT:  PA  (?)  (323,  3). 

.-.  PB  •  PA  =  PT2  (?)  (290).  Q.E.D. 

334.  THEOREM.    If  from  a  fixed  point  without  a  circle  any  secant  be 
drawn,  the  product  of  the  secant  and  its  external  segment  will  be 
constant  for  all  secants. 

Proof  :    Aav  secant  x  ext.  seg.  =  (tan. )2  =  constant  (333). 


162 


PLANE   GEOMETRY 


335.  THEOREM.   If  from  a  fixed  point  without  a  circle  a  secant  and 
a  tangent  be  drawn,  the  tangent  will  be  a  mean  proportional  between 
the  secant  and  its  external  segment. 

Proof :   [Identical  with  proof  of  333;  omitting  the  last  step.] 

336.  THEOREM.    If  from  a  fixed  point  without  a  circle  two  secants 
be  drawn,  these  secants  and  their  external  segments  will  be  reciprocally 
(or  inversely)  proportional. 


Proof :  PB  -  PA  =  PT -  PX  (?)  (334). 
.'.  PB'.PT=PX:PA  (?)  (291). 

Ex.    If  PA  =  3  in.,  and  PB  =  12  in.,  find  the  length  of  PT. 
Ex.  .If  PB  =  21  in.,  PY=  15  in.,  and  PA  =  5  in.,  find  PX. 


Q.B.D. 


337.  THEOREM.  In  any  triangle  the  product  of  two  sides  is  equal  to 
the  diameter  of  the  circumscribed  circle  multiplied  by  the  altitude  upon 
the  third  side. 

Given:     A    ABC-,     circum- 
scribed O  O ;   altitude  BK. 

To  Prove  :  a  •  c  =  d  •  h. 

Proof  :    Draw   chord    CD. 
£  BCD  =  rt.  Z  (?)  (251). 

In  rt.  A  ABX  and  BCD, 
Z  A=Z.D  (?)  (250). 

.*.  these  A  are  similar  (?). 

.'.  e:  <*=A:a(?)  (323,  3). 


Consequently,  a  •  c  =  d  -  h  (?)  (290). 


Q.E.D. 


BOOK   III 


338.  THEOREM.  In  any  triangle  the  product  of  two  sides 
to  the  square  of  the  bisector  of  their  included  angle,  plus  the 
of  the  segments  of  the  third  side  formed  by  the  bisector. 

Given:    A  ABC,    CO   the    bi- 
sector of  Z  A  CB. 

To  Prove  :  a  •  b  =  t2  +  n  •  r. 

Proof:   Circumscribe  a  O    A/ 
about  the  A  ABC. 


163 

is  equal 
product 


\ 


\ 


Produce  CO  to  meet  O  at  D ; 
draw  BD. 

In  A  A  CO  and  BCD,  Z  AGO  = 
</  BCD  (Hyp.). 

And  Z^  =  ZD(?)  (250). 

/.  A  AGO  and  BCD  are  similar  (?)  (314), 

Hence,  b  :  (t  +  x)  =  t  :  a  (?)  (323,  3). 

Therefore,  a  •  5  =  £2  +  *  -  x  (?)  (290). 

CD  and  ^15  are  chords  (Const.).     /.  t  •  x 

Consequently,   a  •  b  =  t2  +  n  -  r  (Ax.  6). 


(329). 

Q.E.D. 


339.   The  projection  of  a  point  upon  a  line  is  the  foot  of  the 
perpendicular  from  the  point  to  the  line 
Thus,  the  projection  of  P  is  J. 


c 


N  M 


The  projection  of  a  definite  line  upon  an  indefinite  line  is 
the  part  of  the  indefinite  line  between  the  feet  of  the  two 
perpendiculars  to  it,  from  the  extremities  of  the  definite  line. 

The  projection  of  AB  is  CD ;   of  US  is  RT-,  of  LM  is  NM. 


164 


PLANE   GEOMETRY 


340.  THEOREM.   If  in  a  right  triangle  a  perpendicular  be  drawn 
from  the  vertex  of  the  right  angle  upon  the  hypotenuse, 

I.    The  triangles  formed  will  be  similar  to  the  given  triangle  and 
similar  to  each  other. 

II.   The    perpendicular  will  be  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

Given:  Rt.   A  ABC-,    CP  JL  c 

to  AB  from  C. 

To  Prove:   I.   A  APC,  ABC, 
and  BPC  similar. 

II.   AP  :  CP  =  CP  :  PB.  #  p  — g 

Proof :   I.   In  rt.  A  APC  and  ABC,  Z  A  =  Z  A  (Iden.). 
/.  A  APC  is  similar  to  A  ABC  (?)   (315). 
In  rt.  A  BPC  and  ABC,  Z  B  =  Z  B  (?). 
.-.  A  BPC  is  similar  to  A  ABC  (?). 

Therefore,  A  APC,  ABC,  and  BPC  are  all  similar  (?)  (320). 
II.    In  the  A  APC  and  BPC,  AP  :  CP  =  CP  :  PB  (?)  (328,  3). 

Q.E.D. 

341.  THEOREM.    If  from  any  point  in  a  circumference  a  perpendicu- 
lar be  drawn  to  a  diameter,  it  will  be  a  mean  proportional  between 
the  segments  of  the  diameter. 

Given:  (?).      To  Prove :  (?). 
Proof  :  Draw  chords  AP  and  BP. 
A  APB  is  a  rt.  A  (?)  (251). 
.'.  AD  :  PD  =  PD  :  DB  (?).  Q.E.D. 

342.  THEOREM.    The  square  of  a  leg  of  a  right  triangle  is  equal  to 
the  product  of  the  hypotenuse  and  the  projection  of  this  leg  upon  the 
hypotenuse.  ~ 

Given:   Rt.  A  ABC;  AC  and 
BC  the  legs. 

To  Prove :  I.  AC2  =  AB  -  AP. 


II.    B<?  =  AB 


BP. 


BOOK  III 


165 


Proof :    I.  The  rt.  A  ABC  and  APC  are  similar  (?)  (340, 1). 
.*.  AB  :AC=AC:AP  (323,  3).      .'.  AC2  =  AB  -  AP  (?). 
II.   Rt.   A  ABC  and  BCP  are  similar    (?). 
.'.  AB  :  BC  =  BC  :  BP  (?).     .'.Isc2  =  AB  -  BP  (?).          Q.E.D. 

Ex.  1.   If,  in  340,  AP  =  3,  P£'  =  27,  find  CP. 

Ex.  2.   If,  in  342,  AP  =  4,  P£  =  21,  find  4C  and  J5C. 

Ex.  3.   If,  in  342,  AB  =  20,  ^  C  =  6,  find  AP,  £P,  CP,  and  BC. 


343.  THEOREM.   The  sum  of  the  squares  of  the  legs  of  a  right  tri- 
angle is  equal  to  the  square  of  the  hypotenuse. 

Given  :  Rt.  A  ABC.    To  Prove  :   AC2  +  5<?  =  Iff. 
Proof :  Draw  CP  _L  to  the  hypotenuse  AB. 
Then  AC*  =  AB  -  AP  (?)  (342). 
And  BC2  =  AB  •  BP  (?).      Adding, 
A(?  +  BC?=AB  •  AP  +  AB-BP  (Ax.  2). 

=  AB  (AP  +  BP)  =  ^B  •  ^4B  =  ij?  (Ax.  4). 
That  is,  Jc2  -I-  5c2  =  ^4jB2.  Q.E.D. 

344.  THEOREM.   The  square  of  either  leg  of  a  right  triangle  is  equal 
to  the  square  of  the  hypotenuse  minus  the  square  of  the  other  leg. 

That  is,    Ic2  =  AB2  -  BC2  ;   and  BC2  =  AB2  -  AC2  (?)  (Ax.  2). 

Ex.    If  AC  =  28  and  BC  =  45,  find  AB. 
Ex.   If  A  C  =  21  and  AB  =  29,  find  BC. 


345.  THEOREM.  In  an  obtuse  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
plus  twice  the  product  of  one  of  these  two  sides  and  the  projection  of  the 
other  side  upon  that  one. 

Given  :  Obtuse  A  ABC ;  etc. 
To  Prove:  c2  =  a2  +  ft2  +  2 ty. 

Proof  :    c2  =  A2  +  Q?  +'  5)2  = 
A2  +  Jt?2+  62+2fy(?)  (343).   . 

But  A2  +  p2  =  a2  (?)  (343). 
,-.  <p  =  a2  +  b*  +  2  6p  (Ax  6). 


c 


M 
Q.E.D. 


166 


PLANE  GEOMETRY 


346.  THEOREM.  In  any  triangle  the  square  of  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
minus  twice  the  product  of  one  of  these  two  sides  and  the  projection  of 
the  other  side  upon  that  one. 

Given:  (?). 

To  Prove :  c2  =  (?}. 


Proof:  £2  = 
+  J0a  +  aa_ 
But, 


?)  (343). 
p  (Ax.  6), 


c 

Q.E.D. 

NOTE.  This  theorem  is  equally  true  in  case  the  triangle  contains  an 
obtuse  angle.  Thus,  in  the  figure  of  345,  suppose  the  projection  of  AB 
is  AM=p.  Then,  a2  =  A2  +  (p  -  6)2  =  A2  +p2  +  62  -  2  ftp  =  etc. 

347.  THEOREM.  I.  The  sum  of  the  squares  of  two  sides  of  a  triangle 
is  equal  to  twice  the  square  of  half  the  third  side  increased  by  twice  the 
square  of  the  median  upon  that  side. 

II.  The  difference  of  the  squares  of  two  sides  of  a  triangle  is  equal 
to  twice  the  product  of  the  third  side  by  the  projection  of  the  median 
upon  that  side. 

Given:  A  ABC;  median  =  c 

m ;  its   projection  =  p ;    and 
side  b  >  side  a. 

To  Prove: 


II.  b*- 

Proof:  In  A  ARC  and  BBC,  AB  =  BE  (Hyp.);  CB  is 
common;  AC  >  BC  (Hyp.).  .-.  Z  ABC  >  Z  BBC  (?)  (87). 
That  is,  Z.  ABC  is  obtuse  and  Z  BBC  is  acute. 

.'.  in  A  ABC,  52  =  (l<?)2  +™?  +  cp  (345), 
and  in  A  BBC,  a2  =  (l<?)2  +  m2  -  cp  (346). 

I.    Adding, 


II.    Subtracting,  b2  -  a2  =  2cp 


(Ax.  2). 
(Ax.  2). 

Q.E.D, 


BOOK  III  167 


348.  If  the  vertices  of  a  triangle  are  denoted  by  A,  B,  (7, 
the  lengths  of  the  sides  opposite  are  denoted  by  a,   6,  0, 
respectively ;    the   altitude    upon   these   sides  by  ha,  hb,  h^ 
respectively ;  the  bisectors  of  the  angles  by  ta,  tb,  tc,  respec- 
tively ;  the  medians  by  wa,  m6,  wc,  respectively  ;  the  segments 
of  the  sides  formed  by  the  bisectors  of  the  opposite  angles 
by  na  and  ra,  nb  and  rb,  nc  and  rc\   and  the  projections  as 
follows :   the  projection  of   side  a  upon  side  5,  by  apb ;  of 
side  a  upon  side  <?,  by  apc\  of  side  b  upon  side  <?,  by  bpc\  etc. 

349.  Formulas.     It  is  assumed  that  a,  6,  c  are  known. 
The  following  values  of  the  various  lines  in  a  triangle  are 
obtained  by  solving  the  equations  already  derived. 

I.  Projections. 

1.  If  Z.  C  is  obtuse,  apb  =  —  — ;  bpa  =  -          — —  ;  etc. 

2.  If  /.  C  is  acute,   apb  =  —          ~  C    \  ipa  =  —         ~  c  ;  etc. 

II.  Altitudes.  hb=   Va2-0jt?62;  ha=  V&2-5jpa2;  etc. 

III.  Medians.  mc  =  \  V2  (a2  +  b2)  -  c2 ;  ma  =  \  V2(62-f  c2)  -a2;  etc. 

IV.  Bisectors.   £c  =  Va&  —  ncrc*  ;  /"„  =  Vfec  —  nara*;  «6  =  Vac  — 
V.  Diameter  of  circumscribed  circle  =  ^;  =^;  =  ^» 

VI.  Largest  Angle. 

1.  Z  C  is  right  if  c2  =  a2  +  b2   (343). 

2.  Z  C  is  obtuse  if  c2  >  a2  +  &2  (345). 

3.  Z  C  is  acute  if  c2<  a2  +  62   (346). 


Ex.  1.  If  the  sides  of  a  triangle  are  a  =  7,  b  =  10,  c  =  12,  find  the 
nature  of  Z  C. 

Ex.  2.  In  the  same  triangle  find  ma.     Find  ?n5.     Find  mc. 

Ex.  3.  In  the  same  triangle  find  apb.  Find  bpa.  Find  aj9c.  Find  t/>c. 

Find  cpa.  Find  Cjo5. 

Ex.  4.  Find  fc«.     Find  As.     Find  Ac. 

Ex.  5.  Find  the  diameter  of  the  circumscribed  circle. 

Ex.  6.  Find  na  and  ra.     Find  n6  and  r&.     Find  nc  and  r,.. 

Ex.  7.  Find  *a.     Find  tb.     Find  fc. 

*  The  segments  n  and  r  can  be  found  by  308  ;  TI&  :  r&  =  c :  a,  etc. 


168  PLANE   GEOMETRY 

CONCERNING  ORIGINALS 

350.  We  should  first  determine  from  the  nature  of  each 
numerical  exercise  upon  which  theorem  it  depends.  By 
applying  the  truth  of  that  theorem,  the  exercise  is  usually 
solved  without  difficulty. 

ORIGINAL    EXERCISES    (NUMERICAL) 

1.  The  legs  of  a  right  triangle  are  12  and  16  inches  ;  find  the  hypote- 
nuse. 

2.  The  side  of  a  square  is  6  feet ;  what  is  the  diagonal? 

3.  The  base  of  an  isosceles  triangle  is  16  and  the  altitude  is  15;  find 
the  equal  sides. 

4.  The  tangent  to  a  circle  from  a  point  is  12  inches  and  the  radius 
of  the  circle  is  5  inches ;  find  the  length  of  the  line  joining  the  point  to  the 
center. 

5.  In  a  circle  whose  radius  is  13  inches,  what  is  the  length  of  a  chord 
5  inches    from   the   center? 

[Draw  chord,  distance,  and  radius  to  its  extremity.] 

6.  The  length  of  a  chord  is  2  feet  and  its  distance  from  the  center 
is  35  inches  ;  find  the  radius  of  the  circle. 

7.  The  hypotenuse  of  a  right  triangle  is  2  feet  2  inches,  and  one  leg 
is  10  inches  ;  find  the  other. 

8.  The  base  of  an  isosceles  triangle  is  90  and  the  equal  sides  are  each 
53 ;  find  the  altitude. 

9.  The  radius  of  a  circle  is  4  feet  7  inches ;  find  the  length  of  the 
tangent  drawn  from  a  point  6  feet  1  inch  from  the  center. 

10.  How  long  is  a  chord  21  yards  from  the  center  of  a  circle  whose 
radius  is  35  yards  ? 

11.  Each  side  of  an  equilateral  triangle  is  4  feet ;  find  the  altitude. 

12.  The  altitude  of  an  equilateral  triangle  is  8  feet ;  find  the  side. 
[Let  x  =  each  side ;  \  x  =  the  base  of  each  rt.  A.] 

13.  Each  side  of  an  isosceles  right  triangle  is  a ;  find  the  hypotenuse. 

14.  If  the  length  of  the  common  chord  of  two  intersecting  circles  is 
16,  and  their  radii  are  10  and  17,  what  is  the  distance  between  their 
centers? 

15.  The  diagonal  of  a  rectangle  is  82  and  one  side  is  80 ;  find  the  other. 

16.  The  length  of  a  tangent  to  a  circle  whose  diameter  is  20,  from  an 
external  point,  is  24.    What  is  the  distance  from  this  point  to  the  center? 


BOOK  III  169 

17.  The  diagonal  of  a  square  is  10 ;  find  each  side. 

18.  Find  the  length  of  a  chord  2  feet  from  the  center  of  a  circle 
whose  diameter  is  5  feet. 

19.  A  flagpole  was  broken  16  feet  from  the  ground,  and  the  top  struck 
the  ground  63  feet  from  the  foot  of  the  pole.     How  long  was  the  pole  ? 

20.  The  top  of  a  ladder  17  feet  long  reaches  a  point  on  a  wall  15  feet 
from  the  ground.     How  far  is  the  lower  end  of  the  ladder  from  the  wall  ? 

21.  A  chord  2  feet  long  is  5  inches  from  the  center  of  a  circle.     How 
far  from  the  center  is  a  chord  10  inches  long?     [Find  the  radius.] 

22.  The  diameters  of  two  concentric  circles  are  1  foot  10  inches  and 
10  feet  2  inches.     Find  the  length  of  a  chord  of  the  larger  which  is  tan- 
gent to  the  less. 

23.  The  lower  ends  of  a  post  and  a  flagpole  are  42  feet  apart ;  the 
post  is  8  feet  high  and  the  pole,  48  feet.     How  far  is  it  from  the  top  of 
one  to  the  top  of  the  other? 

24.  The  radii  of  two  circles  are  8  inches  and  17  inches,  and  their  cen- 
ters are  41  inches  apart.     Find  the  lengths  of  their  common  external 
tangents ;   of  their  common  internal  tangents. 

25.  A  ladder  65  feet  long  stands  in  a  street ;  if  it  inclines  toward  one 
side,  it  will  touch  a  house  at  a  point  16  feet  above  the  pavement ;  if  to 
the  other  side,  it  will  touch  a  house  at  a  point  56  feet  above  the  pave- 
ment.    How  wide  is  the  street  ? 

26.  Two  parallel  chords  of  a  circle  are  4  feet,  and  40  inches  long,  re- 
spectively, and  the  distance  between  them  is  22  inches.     Find  the  radius 
of  the  circle. 

[Draw  the  radii  to  ends  of  chords ;    these  =  hypotenuses  =  R ;   the 
distances  from  the  center  =  x  and  22  —  a;.] 

27.  The  legs  of  an  isosceles  trapezoid  are  each  2  feet  1  inch  long,  and 
one  of  the  bases  is  3  feet  4  inches  longer  than  the  other.     Find  the 
altitude. 

28.  One  of  the  non-parallel  sides  of  a  trapezoid  is  perpendicular  to 
both  bases,  and  is  63  feet  long ;   the  bases  are  41  feet  and  25  feet  long. 
Find  the  length  of  the  remaining  side. 

29.  If  a  =  10,  h  =    6,  find  p,  c,  p',  b. 

30.  If  h  =    S,p'  =    4,  find  6,  c,  p,  a. 

31.  If  a  =  10,  p'  =  15,  find  c,  p,  h,  b. 

32.  If  a  =    9,  b  =  12,  find  c,  p,  p',  h. 

33.  lip  =    3,/  =  12,  find  a,  h,b. 


170  PLANE   GEOMETRY 

34.  The  line  joining  the  midpoint  of  a  chord  to  the  midpoint  of  its 
arc  is  5  inches.     If  the  chord  is  2  feet  long,  what  is  the  diameter? 

35.  If  the  chord  of  an  arc  is  60  and  the  chord  of  its  half  is  34,  what  is 
the  diameter  ? 

36.  The  line  joining  the  midpoint  of  a  chord  to  the  midpoint  of  its 
arc  is  6  inches.     The  chord  of  half  this  arc  is  18  inches.     Find  the 
diameter.     Find  the  length  of  the  original  chord. 

37.  To  a  circle  whose  radius  is  10  inches,  two  tangents  are  drawn 
from  a  point,  each  2  feet  long.     Find  the  length  of  the  chord  joining 
their  points  of  contact. 

38.  The  sides  of  a  triangle  are  6,  9,  11.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  angle. 

39.  Find  the  segments  of  the  longest  side  made  by  the  bisector  of  the 
largest  angle  in  No.  38. 

40.  The  sides  of  a  triangle  are  5,  9,  12.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  exterior  angle.     Also 
of  the  medium  side  made  by  the  bisector  of  its  opposite  exterior  angle. 

41.  In  the  figure  of  306,  if  AC  =  3,  CE  =  5,  EG  =  8,  BD  =  4;  find 
DF  and  FH. 

42.  If  the  sides  of  a  triangle  are  6,  8, 12  and  the  shortest  side  of  a 
similar  triangle  is  15,  find  its  other  sides. 

43.  If  the  homologous  altitudes  of  two  similar  triangles  are  9  and  15 
and  the  base  of  the  former  is  21,  what  is  the  base  of  the  latter? 

44.  In  the  figure  of  324,  AE  =  4=,  EF  =6,  FB  =  9,  GH=15.    Find 
CG  and  CD. 

45.  The  sides  of  a  pentagon  are  5,  6,  8,  9,  18,  and  the  longest  side  of 
a  similar  pentagon  is  78.     Find  the  other  sides. 

46.  A  pair  of  homologous  sides  of  two  similar  polygons  are  9  and  16. 
If  the  perimeter  of  the  first  is  117,  what  is  the  perimeter  of  the  second  ? 

47.  The  perimeters  of  two  similar  polygons  are  72  and  120.     The 
shortest  side  of  the  former  is  4,  what  is  the  shortest  side  of  the  latter  ? 

48.  Two  similar  triangles  have  homologous  bases  20  and  48.     If  the 
altitude  of  the  latter  is  36,  find  the  altitude  of  the  former. 

49.  The  segments  of  a  chord,  made  by  a  second  chord,  are  4  and  27. 
One  segment  of  the  second  chord  is  6,  find  the  other. 

50.  One  of  two  intersecting  chords  is  19  in.  long  and  the  segments  of 
the  other  are  5  in.  and  12  in.     Find  the  segments  of  the  first  chord. 

51.  Two  secants  are  drawn  to  a  circle  from  a  point ;  their  lengths  are 


BOOK  III  171 

15  inches  and  10|  inches.     The  external  segment  of  the  latter  is  10;  find 
the  external  segment  01  the  former. 

52.  The  tangent  to  a  circle  is  1  foot  long  and  the  secant  from  the 
same  point  is  1  foot  6  inches.     Find  the  chord  part  of  the  secant. 

53.  The  internal  segment  of   a  secant  25  inches  long  is  16  inches. 
Find  the  tangent  from  the  same  point  to  the  same  circle. 

54.  Two    secants  to  a  circle  from  a  point   are   1|  feet  and  2  feet 
long ;  the  tangent  from  the  same  point  is  12  inches.     Find  the  external 
segments  of  the  two  secants. 

55.  The  sides  of  a  triangle  are  5,  6,  8.     Is  the  angle  opposite  8  right, 
acute,  or  obtuse  ?    Same  for  the  triangle  8,  7,  4  ? 

56.  The  sides  of  a  triangle  are  8,  9,  12.     Is  the  largest  angle  right, 
acute,  or  obtuse  ?    Same  for  the  triangle  13,  7,  11  ? 

57.  The  sides  of  a  triangle  are  x,  y,  z.     If  z  is  the  greatest  side,  when 
will  the  angle  opposite  be  right?    Obtuse?     Acute? 

58.  The  sides  of  a  triangle  are  6,  8,  9.     Find  the  length  of  the  projec- 
tion of  side  6  upon  side  8 ;  of  side  8  upon  side  9 ;  of  side  9  upon  side  6. 

59.  The  sides  of  a  triangle  are  5,  6,  9.     Find  the  length  of  the  pro- 
jection of  side  6  upon  side  5 ;  of  side  9  upon  side  6. 

60.  Find  the  three  altitudes  in  triangle  9,  10,  17. 

61.  Find  the  three  altitudes  in  triangle  11,  13,  20. 

62.  Find  the  diameter  of  circumscribed  circle  about  triangle  17,  25,  26. 

63.  Find  the  length  of  the  bisector  of  the  least  angle  of  triangle 
7,  15,  20.     Also  of  the  largest  angle. 

64.  Find  the  length  of  the  bisector  of  the  largest  angle  of  triangle 
12,  32,  33 ;  also  of  the  other  angles. 

65.  Find  the  three  medians  in  triangle  4,  7,  9. 

66.  Find  the  product  of  the  segments  of  every  chord  drawn  through 
a  point  4  units  from  the  center  of  a  circle  whose  radius  is  10  units. 

67.  The  bases  of  a  trapezoid  are  12  and  20,  the  altitude  is  8 ;  the 
other  sides  are  produced  to  meet.     Find  the  altitude  of  the  larger  tri- 
angle formed. 

68.  The  shadow  of  a  yardstick  perpendicular  to  the  ground  is  4|  feet. 
Find  the  height  of  a  tree  whose  shadow  at  the  same  time  is  100  yards. 

69.  There  are  two  belt-wheels  3  feet  8  inches  and  1  foot  2  inches  in 
diameter,  respectively.     Their  centers  are  9  feet  5  inches  apart.     Find 
the  length  of  the  belt  suspended  between  the  wheels  if  the  belt  does  not 
cross  itself.     Also  the  length  of  the  belt  if  it  does  cross. 


172  PLANE   GEOMETRY 


SUMMARY 

351.  Triangles  are  proved  similar  by  showing  that  they  have: 

(1)  Two  angles  of  one  equal  to  two  angles  of  the  other. 

(2)  An  acute  angle  of  one  equal  to  an  acute  angle  of  the  other.     [In 
right  triangles.] 

(3)  Homologous  sides  proportional. 

(4)  An  angle  of  one  equal  to  an  angle  of  the  other  and  the  including 
sides  proportional. 

(5)  Their  sides  respectively  parallel  or  perpendicular. 

352.  Four  lines  are  proved  proportional  by  showing  that  they  are : 

(1)  Homologous  sides  of  similar  triangles. 

(2)  Homologous  sides  of  similar  polygons. 

(3)  Homologous  lines  of  similar  figures. 

353.  The  product  of  two  lines  is  proved  equal  to  the  product  of  two 
other  lines,  by  proving  these  four  lines  proportional  and   making  the 
product  of  the  extremes  equal  to  the  product  of  the  means. 

354.  One  line  is  proved  a  mean  proportional  between  two  others  by 
proving  that   two   triangles  which  contain  this   line  in  common  are 
similar,  and  obtaining  the  required  proportion  from  their  sides. 

355.  In  cases  dealing  with  the  square  of  a  line,  one  uses : 

(1)  Similar  triangles  having  this  line  in  common,  or, 

(2)  A  right  triangle  containing  this  line  as  a  part. 

ORIGINAL  EXERCISES   (THEOREMS) 

1.  If  two  transversals  intersect  between  two  parallels,  the  triangles 
formed  are  similar.     [Use  351  (1).] 

2.  Two  isosceles  triangles  are  similar  if  a  base  angle  of  one  is  equal 
to  a  base  angle  of  the  other. 

3.  Two  isosceles  triangles  are  similar  if  the  vertex-angle  of  one  is 
equal  to  the  vertex-angle  of  the  other. 

4.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms  a 
triangle  similar  to  the  original  triangle. 

5.  The  diagonals  of  a  trapezoid  form,  with  the  parallel  sides,  two 
similar  triangles. 


BOOK  III 


173 


6.    Two  circles  are  tangent  externally  at  P ;  through  P  three  lines  are 
drawn,  meeting  one  circumference  in  A,  B,  C] 
and  the   other   in  A',  B',  C1 .     The    triangles 
ABC  and  A'B'C'  are  similar. 


7.  Prove  the  same  theorem  if  the  circles 
are  tangent  internally. 

8.  If  two  circles  are  tangent  externally  at  A' 

P,  and  BB',  CC'  be  drawn  through  P,  terminating  in  the  circumferences, 
the  triangles  PEC  and  PB'C'  will  be  similar. 
[Draw  the  common  tangent  at  P.] 

9.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

10.  If  AD  and  BE  are  two  altitudes  of  triangle 
ABC,  the  triangles  ACD  and  BCE  are  similar. 

11.  Two  altitudes  of  a  triangle  are  reciprocally 
proportional  to  the  bases  to  which  they  are  drawn. 

To  Prove :   AD  :  BE  =  A  C  :  BC. 

12.  The  four  segments  of  the  diagonals  of  a  trapezoid  are  proportional. 

13.  If  at  the  extremities  of  the  hypotenuse  of  a  right 
triangle  perpendiculars  be  erected  meeting  the  legs  pro- 
duced, the  new  triangles  formed  will  be  similar. 

14.  In  the  figure  of  No.  13,  prove : 

(1)  Triangle   A  BC   similar  to  each  of  the  triangles 
ACE  and  BCD. 

(2)  Triangle  ABE  similar  to  triangle  ABD. 

(3)  Triangle  ACE  similar  to  triangle  ABD. 

(4)  Triangle  BCD  similar  to  triangle  ABE. 

(5)  Triangles  ABC,  ABD,  ABE  similar. 

15.  If  AD  and  BE  are  two  altitudes  of  triangle  ABC  (fig.  of  No.  11), 
meeting  at  0,  the  triangles  BOD  and  A  OE  are  similar. 

16.  Triangles  CED  and  ABC   (fig.  of  No.   11)  are 
similar. 

[First  show  A  CAD  and  CEB  similar. 

/.  CA  :  CB  =  CD  :  CE  (?).     Then  use  351  (4).] 

17.  Triangle  ABC  is  inscribed  in  a  circle  and  ^4P  is 
drawn  to  P,  the  midpoint  of  arc  BC,  meeting  chord  CB 
at  D.     The  triangles  ABD  and  A CP  are  similar. 


174 


PLANE   GEOMETRY 


18.  Two   homologous   medians  in  two  similar  triangles  are  in  the 
same  ratio  as  any  two  homologous  sides. 

[Prove  a  pair  of  the  new  triangles  formed,  similar,  by  351  (4).] 

19.  Two  homologous  bisectors  in  two  similar  triangles  are  in  the  same 
ratio  as  any  two  homologous  sides. 

20.  The  radii  of  circles  inscribed  in  two  similar  triangles  are  in  the 
same  ratio  as  any  two  homologous  sides. 

[Bisect  two  pairs  of  homol.  A\  draw  the  altitudes  of  these  new  & ;  etc.] 

21.  The  radii  of  circles  circumscribed  about  two  similar  triangles  are 
in  the  same  ratio  as  any  two  homologous  sides. 

[Erect  J_  bisectors ;  draw  radius  in  each  O.] 

22.  In  any  right  triangle  the  product  of  the  hypotenuse  and  the  al- 
titude upon  it  is  equal  to  the  product  of  the  legs. 

23.  If  two  circles  intersect  at  A  and  B  and  A  C 
and  AD  be  drawn  each  a  tangent  to  one  circle  and 
a  chord  of  the  other,  the  common  chord  AB  will  be 
a  mean  proportional  between  BC  and  BD. 

24.  If  two  circles  are  tangent  externally,  the  chords  formed  by  a 
straight  line  drawn  through  their  point  of  contact  have  the  same  ratio 
as  the  diameters  of  the  circles. 

[Draw  com.  tang,  at  point  of  contact ;  draw  di- 
ameters from  point  of  contact ;  prove  &  sim. ;  etc.] 

25.  If  AB  is  a  diameter  and  BC  a  tangent,  and 
AC  meets  the  circumference  at  D,  the  diameter  is 
a  mean  proportional  between  AC  and  AD. 

[Draw  BD.     Prove  &  containing   AB  similar.] 

26.  If  a  tangent  be  drawn  from  one  extremity  of  a 
diameter,  meeting  secants  from  the  other  extremity, 
these  secants  and  their  internal  segments  will  be  recip- 
rocally proportional. 

To  Prove:    AC: AD  =  ASiAR. 
Proof:    Draw  RS.    In  &  ARS  and  A  CD,  ZA  = 
Z  A  and  /.  ARS  = /.  D.     (Explain.)     Etc. 

27.  If  AB  is  a  chord  and  CE,  another  chord,  drawn 
from  C,  the  midpoint  of  arc  AB,  meeting  chord  AB  at 
D,  A  C  is  a  mean  proportional  between  CD  and  CE. 

Prove  the  above  theorem  and  deduce  that,  CE  •  CD 
is  constant  for  all  positions  of  the  point  E  on  arc  AEB. 


BOOK  in 


175 


28.  If  chord  A  D  be  drawn  from  vertex  A  of  inscribed 
isosceles  triangle  ABC,  cutting  BC  at  E,  AB  will  be 
a  mean  proportional  between  AD  and  AE. 

Prove  the  above  theorem  and  deduce  that,  AD  •  AE 
is  constant  for  all  positions  of  the  point  D  on  arc  BDC. 

29.  If  a  square  be  inscribed  in  a  right  triangle  so 
that  one  vertex  is  on  each  leg  of  the  triangle  and 

the  other  two  vertices  on  the  hypotenuse,  the  side 
of  the  square  will  be  a  mean  proportional  between 
the  other  segments  of  the  hypotenuse. 

To    Prove:   A  D  :DE  =  DE-.EB. 
&ADG  and  BEF  similar. 


First  prove    C       F 


30.  If  the  sides  of  two  triangles  are  respectively  parallel,  the  lines 
joining  homologous  vertices  meet  in  a  point.  (These  lines  to  be  pro- 
duced if  necessary.) 


31.  In  each  of  the  following  triangles,  is  the  greatest  angle  right, 
acute,  or  obtuse,  7,  24,  25  ?    13,  10,  8  ?    19,  13,  23  ? 

32.  Prove  theorem  of  329  by  drawing  two  other  auxiliary  chords. 

33.  Prove  theorem  of  325  if  point  0  is  between  the  parallels. 

34.  Prove  theorem  of  336  by  drawing  A  Y  and  BX. 

35.  In  any  triangle  the  difference  of  the 
squares  of  two  sides  is  equal  to  the  difference  of 
the  squares  of  their  projections  on  the  third  side. 

[AB2  =  (?) ;  BC2  =  (?) .     Subtract,  etc.] 

36.  If  the  altitudes  of  triangle  A  BC  meet  at 
0,AB2-AC2  =  B02-C02. 

[Consult  No.  35  and  substitute.] 

37.  The  square  of  the  altitude  of  an  equilateral  triangle  is  three 
fourths  the  square  of  a  side.     [Let  side  =  a,  etc.] 

38.  If  one  leg  of  a  right  triangle  is  double 
the  other,  its  projection  upon  the  hypotenuse 
is  four  times  the  projection  of  the  other. 

Proof:  (2a)2  =  cjo;  a*  =  cp'  (?). 


176  PLANE   GEOMETRY 

39.  If  the  bisector  of  an  angle  of  a  triangle  bisects  the  opposite  side, 
the  triangle  is  isosceles. 

40.  The  tangents  to  two  intersecting  circles  from  any  point  in  their 
common  chord  produced  are  equal.     [Use  333.] 

41.  If  two  circles  intersect,  their  common  chord,  produced,  bisects 
their  common  tangents.     [Use  333.] 

42.  If  A  B  and  A  C  are  tangents  to  a  circle 
from  A ;    CD  is   perpendicular  to   diameter 
BOX  from  <7;  then  AB-  CD  =  BD.BO. 

[Use  351  (5).] 

43.  If  the  altitude  of  an  equilateral  tri- 
angle is  h,  find  the  side.    [Denote  the  side  by  x  and  half  the  base  by  \  x.~\ 

44.  If  one  side  of  a  triangle  be  divided  by  a  point  into  segments 
which  are  proportional  to  the  other  sides,  a  line  from  this  point  to  the 
opposite  angle  will  bisect  that  angle.     [Converse  of  308.] 

To  Prove :  ^  n  =Z  m  in  fig.  of  308. 

Proof:  Produce  CB  to  P,  making  BP  =  AB;  draw^lP;  etc. 

45.  State  and  prove  the  converse  of  310. 

46.  Two  rhombuses   are  similar  if  an  angle  of  one  is  equal  to  an 
angle  of  the  other. 

47.  If  two  circles  are  tangent  internally  and  any  two  chords  of  the 
greater  be  drawn  from  their  point  of  contact,  they  will  be  divided  propor- 
tionally by  the  circumference  of  the  less. 

[Draw  diameter  to  point  of  contact  and  prove  the  right  A  similar.] 

48.  The  non-parallel  sides  of  a  trapezoid  and  the  line  joining  the  mid- 
points of  the  bases,  if  produced,  meet  at  a  point.    [Use  Ax.  3  and  325.] 

49.  The  diagonals  of  a  trapezoid  and  the  line  joining  the  midpoint 
of  the  bases  meet  at  a  point. 

50.  If  one  chord  bisects  another,  either  segment  of  the  latter  is  a  mean 
proportional  between  the  segments  of  the  other. 

51.  Two  parallelograms  are  similar  if  they  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other  and  the  including  sides  proportional. 

52.  Two  rectangles  are  similar  if  two  adjoining  pairs  of  homologous 
sides  are  proportional. 

53.  If  two  circles   are  tangent  externally, 
the  common  exterior  tangent  is  a  mean  pro- 
portional between  the  diameters. 

[Draw  chords  PA,  PC,  PB,  PD.  Prove,  first, 
A  PD  and  BPC  straight  lines.  Second,  &  ABC 
and  ABD,  similar.] 


BOOK  III  177 

54.  In  any  rhombus  the  sum  of  the  squares  of  the  diagonals  is  equal 
to  the  square  of  half  the  perimeter. 

55.  If  in  an  angle  a  series  of  parallel  lines  be  drawn  having  their  ends 
in  the  sides  of  the  angle,  their  midpoints  will  lie  in  one  straight  line. 

56.  If  ABC  is  an  isosceles  triangle  and  BX  is  the  altitude  upon  AC 
(one  of  the  legs),  BC2  =  2  A C  .  CX.     [Use  346.] 

57.  In  an  isosceles  triangle  the  square  of  one  leg  is 
equal  to  the  square  of  the  line  drawn  from  the  vertex 
to  any  point  of  the  base,  plus  the  product  of  the  seg- 
ments of  the  base. 

Proof  :  Circumscribe  a  O ;  use  method  of  338. 

58.  If  a  line  be  drawn  in  a  trapezoid  parallel  to  the  bases,  the  seg- 
ments  between   the   diagonals    and  the   non- 

parallel  sides  will  be  equal. 

Proof  :   &  A  HI  and  AB C  are  similar  (?);      H/     J^^J       \K 


A  DJK and  DCB also  (?).  :.         =         (?).  

AB      BC  j-  — Q 

We  =  ¥c (?)"    But'if=ff (?)'  (Use  As' 1)!  ete- 

59.  A  line  through  the  point  of  intersection  of  the  diagonals  of  a 
trapezoid,  and  parallel  to  the  bases,  is  bisected  by  that  point. 

60.  If  M  is  the  midpoint  of  hypotenuse  AB  of  right  triangle  ABC, 
AB*  +  EC*  +  AC*  =  8  CM2. 

61.  The  squares  of  the  legs  of  a  right  triangle  have  the  same  ratio  as 
their  projections  upon  the  hypotenuse. 

62.  If  the  diagonals  of  a  quadrilateral  are  perpendicular  to  each  other, 
the  sum  of  the  squares  of  one  pair  of  opposite  sides  is  equal  to  the  sum  of 
the  squares  of  the  other  pair. 

63.  The  sum  of  the  squares  of  the  four  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  of  the  diagonals.     [Use  347,  I.] 

64.  If  DE  be  drawn  parallel  to  the  hypotenuse 
AB  of  right  triangle  ABC,  meeting  A  C  at  D  and 
CB  at  E,  AE2  -f  ~BD2  =  AB2  +  DE2. 

[Use  4  rt.  A  having  vertex  C.] 

65.  If  between  two   parallel  tangents  a  third 
tangent  be  drawn,  the  radius  will  be  a  mean  propor- 
tional between  the  segments  of  the  third  tangent. 

To  Prove  :  BP  :  OP  =  OP  :  PD.      Proof :  A  BOD  is  a  rt.  A  (?).     Etc. 


178 


PLANE  GEOMETRY 


B. 


66.  If  ABCD  is  a  parallelogram,  BD  a  diagonal,  A  G  any  line  from  A 
meeting  BD  at  E,  CD  at  F,  and  BC  (pro- 
duced) at  6r,  ^1,E  is  a  mean  proportional  be- 
tween EF  and  EG. 

Proof  :  &  ABE  and  EZXF  are  similar  (?)  ; 

also  A  ^  DE  and  .BEG  (?).    Obtain  two  ratios  

=  BE:ED  and  then  apply  Ax.  1.  A 

67.  An  interior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  into  segments  proportional  to  the  radii. 

68.  An  exterior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  (externally)  into  segments  proportional  to  the  radii. 

69.  The  common  internal  tangents  of  two  circles  and 
the  common  external  tangents  meet  on  the  line  deter- 
mined by  the  centers  of  the  circles. 

70.  If  from  the  midpoint  P,  of  an  arc  subtended  by 
a  given  chord,  chords  be  drawn  cutting  the  given  chord, 
the  product  of  each  whole  chord  from  P  and  its  segment 
adjacent  to  P  will  be  constant. 

Proof :   Take  two  such  chords,  PA  and  PC;  draw  diameter  PX ;  etc. 
Rt.  &PST  and  PCX  are  similar.  (Explain.) 

71.  If  from  any  point  within  a  triangle  ABC,  perpendiculars  to  the 
sides  be  drawn  —  OR  to  AB,  OS  to  BC,  OT  to  A  C,  AR2  +  BS2  +  ~CT2 
=  BR2  +  CS2  +  AT2.    [Draw  A 0,  BO,  CO.] 

72.  If  two  chords  intersect  within  a  circle  and  at 
right  angles,  the  sum  of  the  squares  of  their  four  seg- 
ments equals  the  square  of  the  diameter. 

To    Prove :     AP2  +  BP*  +  CP2  +  DP2  =  AR*> 
Proof :  Draw  BC,  AD,  RD.     Ch.  BR  is  ±ioAB  (?). 
.*.  CD  is    ||  to  BR    (?).     /.  arc  BC  =  arc  RD  (?). 
Hence,  ch.   BC  =  ch.  RD  (?).     Now,   RD2  =  BC2  =  BP2  +  CP2  (?). 
AD2  =  etc.   (?).    Finally,  AR2  -  AD2  +  RD2  =  etc.  (?). 

73.  The  perpendicular  from  any  point  of  an  arc  upon  its  chord  is  a 
mean    proportional    between    the    perpendiculars 

from  the  same  point  to  the  tangents  at  the  ends 
of  the  chord. 

To  Prove :  PR  :  PT  =  PT :  PS.  Proof :  Prove 
A  ARP  and  BTP  are  aim.,  also  A  APT  and 
PBS  (?).  Thus,  get  two  ratios  each  =  P4  :  PB. 


BOOK  III 


179 


74.  If  lines  be  drawn  from  any  point  in  a  circumfer- 
ence to  the  four  vertices  of  an  inscribed  square,  the  sum 
of  the  squares  of  these  four  lines  will  be  equal  to  twice 
the  square  of  the  diameter. 

Proof:  &  A  PC,  DPB,  are  rt.  A;  etc. 

75.  If  lines  be  drawn  from  any  external  point  to  the 
vertices   of   a  rectangle    ABCD,  the   sum  of  the 
squares  of  two  of  them  which  are  drawn  to  a  pair 

of  opposite  vertices  will  be  equal  to  the  sum  of  the 
squares  of  the  other  two. 

To  Prove  :  PA*  +  PC*  =  PB2  +  PD2. 

Proof :  Draw  PEF  _L  to  the  base,  etc. 

76.  Is  the  theorem  of  No.  75  true  if  the  point 
is  taken  within  the  rectangle  ? 

77.  If  each  of  three  circles  intersects  the  other 
two,  the  three  common  chords  meet  in  a  point. 

Given  :  (?).  To  Prove :  AB,  LM,  RS  meet  at  0. 
Proof:  Suppose  AB  and  LM  meet  at  0.  DrawJ?0 
and  produce  it  to  meet  the  <D  at  X  and  X'.  Prove 
OX  =  OX'  (by  329).  .-.  X,  X',  S  are  coincident. 

78.  In  an  inscribed  quadrilateral  the  sum  of  the 
products  of  the  two  pairs  of  opposite  sides  is  equal  to 
the  product  of  the  diagonals. 

Proof:  Draw  DX  making  Z  CDX  =  Z  AVE; 
&  ADB  and  CDX  are  sim.  (?)  ;  also  A  BCD  and 
ADX  (?).  Hence,  AB  .  DC  =  DB  •  XC  (?),  and 
AD  -BC=DB  •  AX  (?).  Adding  ;  etc. 

79.  If  AB  is  a  diameter,  EC  and  AD  tangents,  meeting  chords  AF 
and  BF  (produced)  at  C  and  D  respectively,  AB  is  a  mean  proportional 
between  the  tangents  EC  and  AD. 

80.  If  ABO  is  isosceles,  AB  =  CO,  and  AO  :  CO  =  CO:  AC,  prove : 

(1)  AB  =  BC=  CO;  (2)  ZABO  =  2ZO-,  (3)  Z  0  =  36°. 
Proofs:    (1)    AO:  AB  =  AB:AC  (Ax.  6).     .'.  A  ABC 

is  sim.  to  A  .4 BO  (?)  (317).    /.  A  ABC  is  isosceles  (?).     Etc. 

(2)  Z   ^J3C=Z   0   (?),  and  Z   CBO  =  Z   0   (?). 
/.  Z  ABO  =  2  Z  0  (Ax.  2). 

(3)  A  0  =  £  the  sum  of  A  of  A  A  OB  (?).     Etc. 

81.  If  from  a  point  A  on  the  circumference  of  a  circle  two  chords  be 
drawn  and  a  line  parallel  to  the  tangent  at  A  meet  them,  the  chords 
and  their  segments  nearer  to  A  will  be  inversely  proportional. 


180  PLANE  GEOMETRY 

CONSTRUCTIONS 

356.  PROBLEM.    To  find  a  fourth  proportional  to  three  given  lines. 
Given  :   Three  lines,  a,  b,  c.    ^  ..-•'& 
Required  :  To  find  a  fourth    f . 

c  ——^—          p..-- 
proportional  to  a,  b,  c.  ...-••'' 

Construction  :  Take  two  in-  ^  ...-< 

definite   lines,    AB    and    AC,       .,.•;:'.".'].  <?. ig \ c 

meeting  at  A.     On  AB  take 

AR=a,  BV=b.     On  AC  take  AS  =  c.     Draw  BS. 

From  V  draw  VW  II  to  RS,  meeting  AC  at  W. 

Statement :  SW  is  the  fourth  proportional  required.    Q.E.F. 

Proof:   In  A  AVW,  BS  is  II  to  VW  (Const.). 

.-.  a  :  b  =  c  :  SW  (?)  (302).  Q.E.D. 

357.  PROBLEM.    To  find  a  third  proportional  to  two  given  lines. 
Given:   (?).  a 

Required:  (?).  y,-X 

.••*'*•• 

Construction :  £•*'*     \ 

Like  that  for  356.  ?/\ 

Statement:  (?).     Proof:  (?).     A"'""£ s — ~ w c 

358.  PROBLEM.    To  divide  a  given  line  into  segments  proportional 
to  any  number  of  given  lines.        ^  I  H      G  B 


Given  :  AB ;  a,  b,  c,  d.  "\/ 

Required:  To  divide  AB 

into   parts   which   shall   be  a D'"-.../ 

proportional  to  a,  b,  c,  d.  k E  *'*'x. 

C— —  **•». 

Construction  :    Draw  AX,  i 

an  indefinite  line,  oblique 
to  AB  from  A.  On  AX  take  AC  =  a,  CD  =  b,  DE  =  c,  EF  =  d. 
Draw  FB.  Then  draw,  through  E,  D,  and  C,  lines  II  to  BF, 
as  EG,  DH,  and  CL 

Statement:  Ai,  in,  HG,  GB  are  the  required  parts.     Q.E.F. 

Proof  :    AI :  a  =  III :  b  =  HG  :  c  =  GB  :  d  (?)  (306).      Q.E  D. 


BOOK   III 


181 


359.   PROBLEM.    To  construct  a  triangle  similar  to  a  given  triangle 
and  having  a  given  side  homologous  to  a  side  of  the  given  triangle. 

c  v  /X 

or/ 


Y, 


A  B   R 

Given  :  A  ABC  and  ES  homologous  to  AB. 
Required :  To  construct  a  A  on  ES  similar  to  A  ABC. 
Construction :  At  E   construct  Z  SEX  =  /.  A ;    at   8  con- 
struct Z  ESY  =  /.  B,  the  sides  of  these  angles  meeting  at  T. 
Statement:   (?).     Proof:   (?)  (314). 

360.   PROBLEM.    To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  given  side  homologous  to  a  side  of  the  given  polygon. 


D1 


A  ^  A1 

Given  :  Polygon  EB ;  line  AfBr  homologous  to  AB. 
Required :  To  construct  a  polygon  upon  A'B',  similar  to 
polygon  EB. 

Construction  :  From  A  draw  diagonals  AC  and  AD. 
On  A'B'  construct  A  A'B'C'  similar  to  A  ABC  (by  359). 
On  A'c'  construct  A  A'C'D'  similar  to  A  ACD.    Etc. 

Statement:  (?). 

Proof:  (?)  (328). 


182 


PLANE   GEOMETRY 


:/  a 

S 

"\ 

b 

\ 

\.\ 

\  "E 
a  

b 

1 

o 

c 

361.  PROBLEM.   To  find  the  mean  proportional  between  two  given 
lines.  ^ 

Given :  Lines  a  and  b. 

Required :  To  find  the 
mean  proportional  between 
them. 

Construction  :  On  an  in- 
definite line,  AX,  take  AB 
—  a  and  BC=  b.  Using  o,  the  midpoint  of  AC,  as  center,  and 
AO  as  radius,  describe  the  semicircumference,  ADC.  At  B 
erect  BD  _L  to  AC,  meeting  the  arc  at  D.  Draw  AD  and  CD. 

Statement:  BD  is  the  mean  proportional  required.     Q.E.F. 

Proof :  a  :  BD  =  BD  :  b  (?)  (341).  Q.E.D. 

362.  A  line  is  divided  in  extreme  and  mean  ratio  if  one 
segment  is  a  mean  proportional  between  the  whole  line  and 
the  other  segment.      In  other  words,  if  a  line  is  to  one  of  its 
parts,  as  that  part  is  to  the  other  part,  the  line  is  divided 
in  extreme  and  mean  ratio. 


363.   PROBLEM.    To  divide  a  line  into  extreme  and  mean  ratio. 
Given :  Line  AB  =  a. 

R 

Required :  To  divide 
AB  into  extreme  and 
mean  ratio,  that  is,  so 
that  AB:AF=AF:FB. 

Construction :  At  B 
erect  BR,  _L  to  AB  and 
=  AB.  Using  c,  the 
midpoint  of  BR,  as  cen-  ...•••'*' 


\E 


a-x 


ter,  and  CB  as  radius,    --••'''       * 

describe  a  O.      Draw 

AC  meeting  O  at  D  and  E.    On  AB  take  AF=  AD ;  let  AF=  x. 

Statement :  .F7 divides  AB  so  that  AB  :  AF=  AF :  FB.   Q.E.F. 


BOOK  in  183 

Proof :  DE=  a  (?)  (203).     .-.  AE  =  a  +  x  (Ax.  4). 
AB  is  tangent  to  the  O   (?)  (215).' 

.'.  AE-  AD=AB2  (?)  (333). 
.*.  («  +  #)  x  —  a2 (Ax.  6). 
That  is,  x2  =  a2  —  a#,  or  #2  =  a  (a  —  a?). 

.'.a:x  =  x:a  —  x  (?)  (291). 
That  is,   ^LB  :  AF  =  AF:FB  (Ax.  6).  Q.E.D. 

364.  PROBLEM.    To  divide  a  line  externally  into  extreme  and  mean 
ratio. 

Given:  (?).  .......* 

Required:  (?). 

Construction:  The  same  as  in  / 

363,  except  AFr  is  taken  on  BA  I  ...,.£' 

produced,  =  AE.  \    .....-"""  / 


.-- 


' 


Statement :  ^JB  :  AFr  =  AFr :  BF'.  Q.B.F. 

Proof:  AB  is  tangent  to  the  O  (?).     AB  =  BB  =  DE  (?). 
.'.  AE:AB  =  AB:  AD  (?)  (335). 

Hence,  ^^  +  ^15  :  ^^  ==  AB  +  ^4-D  :  AB  (?)  (294). 

But  ^1^  +  AB  =  BFf,  and  ^4JB  +  AD  =  AE=  AFr  (Const.). 

/.  BF'  :  AFf  =  AF'  :  AB  (Ax.  6). 

That  is,  AB  :  AFr  ==  AFf  :  BFf.  Q.E.D. 

365.   The  lengths  of  the  several  lines  of  363  and  364  may 
be  found  by  algebra,  if  the  length  of  AB  is  known. 
Thus,  if  AB  =  a,  we  know  in  363,  a  :  x  =  x  :  a  —  x. 
Hence,  x2  =  a2  —  ax.      Solving  this  quadratic, 
x=AF=  -|-a(V5  —  1)  ;  also,   a—  x  —  BF  =  \a  (3—  V5). 
Likewise,  in  364,  if  AB  =  a,  AFf  =  y,  a\y —  y  \a-\-y. 
Solving  for  y,    y  =  AF!  =  |  a  (V§  + 1). 
Also  a  +  y  =  BF'  =  \a  (3  +  V5). 


184  PLANE   GEOMETRY 


ORIGINAL   CONSTRUCTIONS 

1.  To  construct  a  fourth  proportional  to  lines  that  are  exactly  3  in., 

5  in.,  and  6  in.  long.     How  long  should  this  constructed  line  be  ? 

2.  To  construct  a  mean  proportional  between  lines  that  are  exactly 
4  in.  and  9  in.     How  long  should  this  constructed  line  be  ? 

3.  Will  a  fourth  proportional  to  three  lines  5  in.,  8  in.,  and  10  in.,  be 
the  same  length  as  a  fourth  proportional  to  5  in.,  10  in.,  and  8  in.  ?    to  8 
in.,  10  in.,  and  5  in.?  to  10  in.,  5  in.,  and  8  in.  ? 

4.  To  construct  a  third  proportional  to  lines  that  are  exactly  3  in.  and 

6  in.  long. 

5.  To  produce  a  given  line  AB  to  point  P,  such  that  AB :  AP  =  3  :  5. 
[Divide  A  B  into  three  equal  parts,  etc.] 

6.  To  divide  a  line  8  in.  long  into  two  parts  in  the  ratio  of  5:7. 
[Divide  the  given  line  into  12  equal  parts.] 

7.  To  solve  No.  6  by  constructing  a  triangle.     [See  308.] 

8.  To  divide  one  side  of  a  triangle  into  segments  proportional  to  the 
other  two  sides. 

9.  To  divide  one  side  of  a  triangle  externally  into  segments  propor- 
tional to  the  other  sides. 

10.  To  construct  two  straight  lines  having  given  their  sum  and  ratio. 
[Consult  No.  6.] 

11.  To  construct  two  straight  lines  having  given  their  difference  and 
ratio.     [Consult  No.  5.] 

12.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  perimeter.     [First,  use  358.] 

13.  To  construct  a  right  triangle  having  given  its  perimeter  and  an 
acute  angle.     [Constr.  a  rt.  A  having  the  given  acute  Z.    Etc.] 

14.  To  construct   a   triangle   having  given    its  perimeter  and  two 
angles.     [Constr.  a  A  having  the  two  given  A.    Etc.] 

15.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  altitude. 

16.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  base. 

17.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  perimeter. 

18.  To  construct  a  parallelogram  similar  to   a  given  parallelogram 
and  having  a  given  base. 


BOOK  III 


185 


19.  To  construct  a  parallelogram  similar  to  a  given  parallelogram 
and  having  a  given  perimeter. 

20.  To  construct  a  parallelogram  similar  to  a  given   parallelogram 
and  having  a  given  altitude. 

21.  Three  lines  meet  at  a  point ;  it  is  required  to  draw  through  a 
given  point   another  line,    which    will    be 

terminated  by  the  outer  two  and  be  bisected 
by  the  inner  one. 

Construction :  From  E  on  BD  draw  Us. 
Etc.  Through  P  draw  R T  \\  to  GF.  v^-tKX XT  C 

Statement :    RS  =  ST.  B          F 

22.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a  given  triangle. 
Construction  :   Circumscribe  a  O  about  the  given  A ;  draw  radii  to  the 

vertices ;  at  center  of  given  O  construct  3  A  =  to  the  other  central  angles. 

23.  To  circumscribe  about  a  given  circle  a  triangle  similar  to  a  given 
triangle. 

Construction :   First,  inscribe  a  A  similar  to  the  given  A. 

24.  To  construct  a  right  triangle,  having  given  one  leg  and  its  projec- 
tion upon  the  hypotenuse. 

25.  To  inscribe  a  square  in  a  given  semicircle. 

Construction :    At  B  erect  BD  _L  to  AB  and  =  AB ;  draw  DC,  meeting 
Oat  R;  draw  RU  \\toBD.  Etc.     Statement:   (?). 
Proof:   RSTU  is  a  rectangle  (?). 
A  CRU  is  similar  to  A  CBD  (?). 
.'. CU :  CB  =  R U :  BD  (?).   But  CB=\BD  (?). 
/.  CU=\RU(^).  Etc. 

26.  To  inscribe  in   a  given  semicircle,  a  rec- 
tangle similar  to  a  given  rectangle. 

Construction  :  From  the  midpoint  of  the  base    *      T    C 
draw  line  to  one  of  the  opposite  vertices.     At  given  center  construct  an  Z 
=  the  Z  at  the  midpoint.     Proceed  as  in  No.  25. 

Proof  :   First,  prove  a  pair  of  A  similar.   Etc. 

27.  To  inscribe  a  square  in  a  given  triangle.  A  P 
Construction:    Draw  altitude    AD]    con- 
struct the  square  A  DEF  upon  AD  as  a  side  ; 

draw  BF  meeting  A  C  at  R. 

Draw  RU  \\  to  AD;  RS  \\  to  BC.  Etc. 

Statement:     (?).      Proof:     &  BRU   and 
BFE  are  similar  (?).    Also  &  BRS  and  BAF  (?). 

Thence  show  that  SR  =  RU.   Etc. 


B 


\ 


DU  C 


{ 

i 
i 


186  PLANE   GEOMETRY 

28.  To  inscribe  in  a  given  triangle  a  rectangle  similar   to   a  given 
rectangle. 

Construction :    Draw   the    altitude.     On  this  construct  a   rectangle 
similar  to  the  given  rectangle. 
Proceed  as  in  No.  27. 

29.  To  construct  a  circle  which  shall  pass 
through  two  given  points  and  touch   a  given 
line. 

Given  :   Points  A  and  B  ;  line  CD. 

r      t  .••  ^^     •     ^  Q 

Construction:   Draw  line  AB  meeting   CD  c-"     P 
at  P.     Construct  a  mean  proportional  between  PA  and  PB   (by 361). 
On  PD  take  PR  =  this  mean.     Erect  OR  JL  to  CD  at  R,  meeting  JL  bi- 
sector of  AB  at  O.  '  Use  O  as  center,  etc. 

30.  To  construct  a  line  =  V2  in.    [Diag.  of  square  whose  side  is  1  in.] 

31.  To  construct  a  line  =  V^Tin. 

[Hyp.  of  a  rt.  A,  whose  legs  are  1  in.  and  2  in.  respectively.] 

32.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  V2". 

33.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  V5. 

34.  To  construct  a  line  as,    if  x  =  — ,  and  a,  b,  c  are  lengths  of  three 
given  lines.     [That  is,  to  construct  x,  if  c  :  a  =  b  :  x  (291).] 

35.  To  construct  a  line  x,ifx  =  —.     [3c :  a  =  b  :  x.'] 

oc 

36.  To  construct  a  line  x,  if  x  —  Vab.     [a  :  x  =  x :  &.] 

37.  To  construct  a  line  a:,  if  x  =  — . 

c 

38.  To  construct  a  line  a?,  if  a;  =  Vo^—  62.     [a  +  6 :  a;  =  x :  a  —  6.] 

39.  To  construct  a  line  x,  if  *  =  — . 

c 

40.  To  construct  a  line  y,  if  ay  =  f  &2. 

41.  To  construct  a  line  =  VI()  in. 

42.  To  construct  a  line  =  2  V6  in. 

43.  To  construct  a  line  =  Va2  +  b2,  if  a  and  b  are  given  lines. 


UNIT   OF   LENGTH 


BOOK   IV 

AREAS 

366.  The  unit  of  surface  is  a  square  whose 
sides  are  each  a  unit  of  length. 

Familiar  units  of   surface   are  the  square  inch,  the 
square  foot,  the  square  meter,  etc. 

367.  The  area  of  a  surface  is  the  number  of 
units  of  surface  it  contains.     The  area  of  a  sur- 
face is  the  ratio  of  that  surface  to  the  unit  of 
surface. 

Equivalent  (=0=)  figures  are  figures  having  equal  areas. 

NOTE.  It  is  often  convenient  to  speak  of  "triangle,"  "rectangle," 
etc.,  when  one  really  means  "  the  area  of  a  triangle,"  or  "  the  area  of  a 
rectangle,"  etc. 

368.  THEOREM.   If  two  rectangles  have  equal  altitudes,  they  are  to 
each  other  as  their  bases. 

Given :  Rectangles  AC  and   D  c    H 

EG   having  =  altitudes,    and 
their  bases  being  AB  and  EF. 

To  Prove : 


AC  '.  EG  —  AB:  EF. 


J     E 


Proof:  I.  If  A  B  and  EF  are  commensurable. 

There  exists  a  common  unit  of  measure  of  AB  and  EF 
(238).  Suppose  this  unit  is  contained  3  times  in  AB  and 
5  times  in  EF.  Hence,  AB  :  EF=  3  :  5  (Ax.  3). 

Draw  lines  through  these  points  of  division,  _L  to  the  bases. 
These  will  divide  rectangle  AC  into  three  parts  and  EG  into 
5  parts,  and  all  of  these  eight  parts  are  equal  (?)  (140). 

Hence,  AC  :  EG=  3  :  5  (?). 

.'.  AC  :  EG  =  AB  :  EF  (Ax.  1).  Q.E.D. 

187 


188 


PLANE   GEOMETRY 


C    H 


SG 


II.  If  AB  and  EF  are  incom- 
mensurable. 

There  does  not  exist  a  com- 
mon unit  (?)  (238).  Divide 
AB  into  several  equal  parts. 
Apply  one  of  these  as  a  unit 
of  measure  to  EF.  There  will  be  a  remainder,  EF  left 

over  (Hyp.).     Draw  E8  J.  to  EF.     Now,  —  =  ~  (Case  I). 

E8       EE 

Indefinitely  increase  the  number  of  equal  parts  of  AB; 
that  is,  indefinitely  decrease  each  part,  or  the  unit  or 
divisor.  Hence  the  remainder,  EF,  will  be  indefinitely  de- 
creased (?). 

That  is,  EF  will  approach  zero  as  a  limit, 
and,  EFGS  will  approach  zero  as  a  limit. 
Hence,  EE  will  approach  EF  as  a  limit  (?), 
and   ES  will  approach  EG  as  a  limit  (?). 


AC 


AC 


Therefore,  -  -  will  approach  -  -  as  a  limit  (?), 

ES  EG 

and  —  will  approach  —  as  a  limit  (?). 
EE  EF 

.-.—  =  —  (?)   (242). 

EG       EF 


Q.B.D, 


369.  THEOREM.    Two  rectangles  having  equal  bases  are  to  each 
other  as  their  altitudes.    (Explain.) 

370.  THEOREM.    Any  two  rectangles  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes. 


G 

X 

B 

A 

1  ] 

Given :  Rectangles  A  and  B  whose  altitudes  are  a  and  o 
and  bases  b  and  d  respectively. 


BOOK  IV 


189 


To  Prove  :  A  :  B  =  a  -  b  :  c  •  d. 

Proof  :  Construct  a  third  rectangle  X  whose  base  is  b  and 
whose  altitude  is  c. 

Then  A  :  X  =  a  :  c  (?)  (369).    Also,  X  :  B  =  b  :  d  (?). 
Multiplying,  A  :  B  =  a  •  b  :  c  •  d  (?)  (Ax.  3).  Q.E.D. 

371.   THEOREM.    The  area  of  a  rectangle  is  equal  to  the  product  of 
its  base  by  its  altitude. 

Given  :  Rectangle  E,  whose 
base  is  b  and  altitude,  h. 

To  Prove  :  Area  of  K  =  b  -  h. 

Proof  :  Draw  a  square  Z7,  each 
of  whose  sides  is  a  unit  of 
length.  This  square  is  a  unit  of  surface  (366). 


:  '  " :" 

1!   U   j 

: : 

1 


Now,-  =  —  =  b-h  (370).     But  -=  area  of 


(367). 
Q.E.D. 


.-.  area  of  E  —  b  •  h  (Ax.  1). 

372.  THEOREM.   The  area  of  a  square  is  equal  to  the  square  of 
its  side.    (See  371.) 

373.  THEOREM.   The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Given  :  O  ABCD  whose  base  is  b  and  altitude,  h. 

To  Prove  :  Area  of  ABCD  =  b-h. 

Proof:  From  A  and  5,  the  extremities  of  the  base,  draw 
Js  to  the  upper  base  meeting   F         D  EC 

it  in  F  and  E  respectively. 

In  rt.  A  ADF  and  BCE, 
AF  =  BE  (?),  AD=BC  (?). 
.'.  A  ADF  =  A  BCE  (?). 


Now,  from  the  whole  figure  A  B 

take  A  ADF  and  parallelogram  ABCD  remains  ;  and  from  the 
whole  figure  take  A  BCE  and  rectangle  ABEF  remains. 

.'.  O  ABCD  o  rect.  ABEF  (Ax.  2). 

Rect.  ABEF  =  b  •  h  (?).     .'.E]ABCD  =  b-  A  (Ax.  1).   Q.E.D. 


190  PLANE  GEOMETRY 

374.  COR.    All  parallelograms  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

375.  THEOREM.    Two  parallelograms  having  equal  altitudes  are  to 
each  other  as  their  bases. 

Proof:  P  =  b.h',  Pf  =  b'.h(?). 


376.  THEOREM.   Two  parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes. 

Proof:   (?). 

377.  THEOREM.   Any  two  parallelograms  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes. 

Proof:   (?). 

378.  THEOREM.   The  area  of  a  triangle  is  equal  to  half  the  product 
of  its  base  by  its  altitude. 

Given  :  A  ABC;  base  =  b  ;    R. 
altitude  =  h. 

To  Prove  : 

Area  of  A  ABC  =  £  b  •  h. 
Proof  :    Through  A  draw 
AB  II  to  BC  and  through  C 

draw  CB  II  to  AB,  meeting  AB  at  B.    Now  ABCR  is  a  £U  (?). 
Area  O  ABCR  =b  •  h  (?).      |  O  ABCB  =  J  b  -  h    (Ax.  3). 
AlsoAABC=lOABCR(?)  (132). 

.*.  AABC=  %b  -  h  (Ax.  1).  Q.B.D. 

379.  COR.   If  a  parallelogram  and  a  triangle  have  the  same  base 
and  altitude,  the  triangle  is  equivalent  to  half  the  parallelogram. 

380.  COR.   All  triangles  having  equal  bases  and  equal  altitudes  are 
equivalent. 

381.  COR.   All  triangles  having  the  same  base  and  whose  vertices 
are  in  a  line  parallel  to  the  base  are  equivalent  (?). 


BOOK  IV  191 

382.  THEOREM.   Two  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

Proof  :   A  21  =  J  6  •  A  ;  A  T'  =  J  b'  h  (?). 

A  T  _^bh  _b  ^ 
'  A  Tf  ~  %b'h~~  bf 

383.  THEOREM.  Two  triangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Proof:   (?). 

384.  THEOREM.    Any  two  triangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 

385.  THEOREM.   The  area  of  a  right  triangle  is  equal  to  half  the 
product  of  the  legs. 

386.  THEOREM.   The  area  of  a  trapezoid  is  equal  to  half  the  product 
of  the  altitude  by  the  sum  of  the  bases. 

Given:  Trapezoid    ABCD; 
altitude  =  h ;  bases  =  b  and*?. 

To  Prove : 
Area  ABCD  =  ±  h  -  (b  +  c). 

Proof  :  Draw  diagonal  AC. 
Now  consider  the  A  ABC  and 
ADC  as  having  the  same  altitude,  A,  and  their  bases  b  and  c?, 
respectively. 

Now,       A  ABC  =  |  b  -  h  (?), 

and  AADC=  ^c-  A  (?). 

Adding,       A  ABC  +  AADC  =  \b-  A-f  \c-~k     (Ax.  2). 
That  is,       trapezoid  ABCD  =  J  A  •  (b  +  c*)     (Ax.  6).      Q.B.D. 

387.  THEOREM.    The  area  of  a  trapezoid  is  equal  to  the  product  of 
the  altitude  by  the  median. 

Proof :   Area  ABCD  =  J  A  •  (b  -f  c)  =  A  •  J  (b  +  e). 

But  1  (b  +  c)  =  median  (144). 

Hence,  area  of  trapezoid  ABCD  =  A  •  m    (Ax.  6).       Q.E.D. 


192  PLANE   GEOMETRY 

388.  THEOREM.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  they  are  to  each  other  as  the  products  of  the  sides 
including  the  equal  angles. 

Given  :  A  ABC  and  DEF, 
Z.A  =  Z.D. 

To  Prove  : 

A  ABC  _  AB  -  AC 

A  DEF      DE  -  DF 

Proof  :  Superpose  A  ABC 
upon  A  DEF  so  that  the  equal 
A  coincide  and  BC  takes  the 
position  denoted  by  GH.  Draw  GF. 

Now  A  DGH  and  DGF  have  the  same  altitude  (a  _L  from  G 
to  DF),  and  A  DGF  and  DEF  have  the  same  altitude  (a  _L 
from  Fto  DE). 

=  52(?)  (382).     And 


A  DGF      DF  A  DEF      DE 

TVT    14.-    i    •  A  DGH      DH  -  DG  .<)\ 

Multiplying,         -_=__(?). 

That  is,  AABC^AB-AC       ^  g> 

A  D^F      DE  -  DF  Q.E.D. 


Ex.  1.  Prove  theorem  of  386  by  drawing  through  C  a  line  parallel  to 
^4Z),  dividing  the  trapezoid  into  a  parallelogram  and  a  triangle. 

Ex.2.  Which  includes  the  other,  the  word  "equal"  or  the  word 
"equivalent"?  Which  of  these  words  conveys  no  idea  of  shape? 

Ex.  3.  What  is  the  area  of  a  parallelogram  whose  base  is  8  inches 
and  altitude  is  5  inches  ?  What  is  the  area  of  a  triangle  having  the  same 
base  and  altitude  ? 

Ex.  4.  Is  the  area  of  a  triangle  equal  to  half  the  base  multiplied  by 
the  whole  altitude?  Or  half  the  altitude  multiplied  by  the  whole  base? 
Or  half  the  base  multiplied  by  half  the  altitude?  Or  half  the  product 
of  the  base  by  the  altitude  ? 

Ex.  5.  If,  in  the  figure  of  388,  one  triangle  is  four  times  as  large  as 
the  other,  AB  =  10,  A  C  =  6,  DE  =  16,  find  DF. 

Ex.  6.  The  base  of  a  triangle  is  20  and  its  altitude  is  15.  The  bases 
of  an  equivalent  trapezoid  are  13  and  11;  find  its  altitude. 


BOOK  IV 


193 


389.   THEOREM.   Two    similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 


Given  :   Similar  A  ABC  and  DEF. 
TO  Prove  : 


A  DEF 

Proof:  ONE  METHOD.  Z  BAC  =  /.EDF  (?)  (323,1). 

i\  ABC       AB  '  AC/ 


/-o-v    /^OQQN 

(r)  (doo). 


=  4*  x  ^.     Now,  ^  =  ^  (?). 
A  DEF      DE      DF  DF       DE 


A  DEF       DE  •  DF 


That  is, 


x       =      -  (Ax.  6). 

A  DEF      DE      DE 


DE      DF       EF  * 

AABC_  AJ?  __  AC*  _  BC2 
-^-—  a-—  2 


=         =  (297). 

up*       EF 


Q.B.D. 


ANOTHER  METHOD.    Denote  a  pair  of  homologous  alti 
tudes  by  h  and  A',  and  the  corresponding  bases  by  b  and  6'. 


But  - 


That  is, 


A  DEF 


(822) 


Q.E.D. 


194 


PLANE   GEOMETRY 


390.  THEOREM.   Two  similar  polygons  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 

B' 


E  D  E' 

Given :  Similar  polygons  ABODE  and  A'B'C'D'E'. 
To  Prove :  ABODE  :  A'B'C'D'E'  =  Ze2 :  A'B'*  =  etc. 

Proof:  Draw  from  homologous  vertices,  A  and  Af,  all  the 
pairs  of  homologous  diagonals,  dividing  the  polygons  into  A. 
These  A  are  similar,  in  pairs  (?)  (327). 

AR         AB2        ,          -     AS         ~CD2          AB2 


'A*'      A 

75*  v            A.'    W     Z5" 

A  r  _  ^2  _  is2  ,9. 

AT'       JxP      A>B>^ 

Therefore, 
Hence 

AR_    A  8  _  AT   x^x    ^ 
A  Rf  ~  A  $'  ~  A  Tf 
?  +AS   +AT        AR   (^  r3()n 

But 

.   polygon 

t'  +  As'  +  AT*      AR' 

ABODE       _  AB2  __  BO2 

polygon 

A'B'C'D'E'       jr#*      ^,2 

Ex.  1.    The  base  of  a  triangle  is  6.     Find  the  base  of  a  similar  tri- 
angle that  is  9  times  as  large.     Five  times  as  large. 

Ex.  2.   The  area  of  a  polygon  is  104  and  its  longest  side  is  12.     What 
is  the  area  of  a  similar  polygon  whose  longest  side  is  15  ? 


BOOK  IV 


195 


391.  THEOREM.  The  square  described  upon   the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares  described  upon 
the  legs. 

Given:    (?). 

To  Prove:  (?). 

Proof  :  Draw  CL  J_  to 
AB,  meeting  AB  at  K  and 
ED  at  L.  Draw  BF  and  CE. 

Now,  AACB,  ACG,  and 
BCH  are  all  rt.  A  (?). 

Hence,  ACH  and  BCG 
are  straight  lines  (?)  (45). 

Also,  AELK  and  BDLK 
are  rectangles  (?)  (127). 

In  A  ABF  and  ACE, 
AB=AE,  AF—AC  (128), 
and  Z  BAF  =  /_  CAE.  (Each 
=  a  rt.Z  +  /.BAG.) 

.'.  A  ABF  =  A  ACE  (?)  (52). 

Also,  A  ABF  and  square  AG  have  the  same  base,  AF,  and 
the  same  altitude,  AC. 

Hence,    square   AG  ^=2  AABF  (379). 

Similarly,  rectangle  AKLE o2AACE(?). 

Therefore,  rectangle  AKLE  =0=  square  ACGF  (Ax.  1). 

By  drawing  AI  and  CD,  it  is  proved  in  like  manner,  that 
rectangle  BDLK  =c=  square  BCHI.     Then,  by  adding, 
square  ABDE  =0=  square  ACGF-\-  square  BCHI  (Ax.  2).    Q.E.D. 

392.  THEOREM.  The  square  described  upon  one  of  the  legs  of  a  right 
triangle  is  equivalent  to  the  square  described  upon  the  hypotenuse 
minus  the  square  described  upon  the  other  leg.    (Explain.) 


Ex.  1.  If  the  legs  of  a  right  triangle  are  15  and  20,  what  is  the  hypote- 
nuse ?  If  the  legs  are  ?n2  -  1  and  2 m,  what  is  the  hypotenuse? 

Ex.  2.  What  is  the  difference  in  the  wording  of  the  theorems  of  343 
and  391  ?  Which  proof  is  purely  algebraic  ?  Which  is  geometric  ? 


196 


PLANE  GEOMETRY 


393.  THEOREM,  ff  the  three  sides 
of  a  right  triangle  are  the  homolo- 
gous sides  of  three  similar  polygons, 
the  polygon  described  upon  the  hypote- 
nuse is  equivalent  to  the  sum  of  the 
two  polygons  described  upon  the  legs. 

Proof :   -  =  =  (?). 

%        AB2 


Adding, 


8  +  T       AC    +  BC 


=  £"     •    ""  =  gg-  =  l.     (Explain.) 


And 


.'.  E  =c=  S  +  T  (?).  Q.E.D. 

394.  COR.   If  the  three  sides  of  a  right  triangle  are  the  homolo- 
gous sides  of  three  similar  polygons,  the  polygon  described  upon  one 
of  the  legs  is  equivalent  to  the  polygon  described  upon  the  hypote- 
nuse minus  the  polygon  described  upon  the  other  leg. 

395.  THEOREM.      The    two 
squares  described  upon  the  legs 
of  a  right  triangle  are  to  each 
other  as  the  projections  of  the 
legs  upon  the  hypotenuse. 


Proof  :    Scluare  8  =  ^L 
Square   T 


AB 


AB 


AP      AP        sv      ^    - 
—  =  --     (Explain. 
BP       BP 


396.  THEOREM.  If  two  similar  poly- 
gons are  described  upon  the  legs  of  a 
right  triangle  as  homologous  sides,  they 
are  to  each  other  as  the  projections  of 
the  legs  upon  the  hypotenuse. 

2 

Proof:   £«,4Lii4f.   (Explain.) 
T     ~~(         BP 


P        0 


BOOK  IV  197 


ORIGINAL   EXERCISES   (THEOREMS) 

1.   If  one  parallelogram  has  half  the  base  and  the  same  altitude  as 
another,  the  area  of  the  first  is  half  the  area  of  the  second. 

2    If  one  parallelogram   has  half  the  base  and  half  the  altitude  of 
another,  its  area  is  one  fourth  the  area  of  the  second. 

3.  State  and  prove  two  analogous  theorems  about  triangles. 

4.  If  a  triangle  has  half  the  base  and  half  the  altitude  of  a  paral- 
lelogram, the  triangle  is  one  eighth  of  the  parallelogram. 

5.  The  area  of  a  rhombus  is  equal  to  half  the  product  of  its  diagonals. 

6.  The  diagonals  of  a  parallelogram  divide  it  into  four  equivalent 
triangles. 

7.  The  diagonals  of  a  trapezoid  divide  it  into  four  triangles,  two  of 
which  are  similar  and  the  other  two  are  equivalent. 

8.  If  a  parallelogram  has  half  the  base  and  half  the  altitude  of  a 
triangle,  its  area  is  half  the  area  of  the  triangle. 

9.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms  a 
triangle  whose  area  is  one  fourth  the  area  of  the  original  triangle. 

10.  The  line  joining  the  midpoints  of  two  adjacent  sides  of  a  paral- 
lelogram cuts  off  a  triangle  whose  area  is  one  eighth  of  the  area  of  the 
parallelogram. 

11.  If  one  diagonal  of  a  quadrilateral  bisects 
the  other,  it  also  divides  the  quadrilateral  into 
two  equivalent  triangles. 

To  Prove :  A  ABC  -  A  ADC. 

12.  Either  diagonal  of  a  trapezoid  divides  the  figure  into  two  triangles 
whose  ratio  is  equal  to  the  ratio  of  the  bases  of  the  trapezoid.     Prove 
two  ways.     [By  382  and  by  388.] 

13.  If,  in  triangle  ABC,  D  and  E  are  the  midpoints  of  sides  AB  and 
A C  respectively,  £BCD^&BEC. 

14.  If  the  diagonals  of  quadrilateral  A  BCD  meet  at  E  and  A  ABE  =0= 
A  CDE,  the  sides  A  D  and  EC  are  parallel.     [Prove  A  ABD  -  A  A  CD.'] 

15.  The  square  described  upon  the  hypotenuse  of  an  isosceles  right 
triangle  is  equivalent  to  four  times  the  triangle. 

16.  The  square  described  upon  the  diagonal  of  a  square  is  double  the 
original  square. 


198 


PLANE   GEOMETRY 


17.  Any  two  sides  of  a  triangle  are  reciprocally  proportional  to  the 
altitudes  upon  them.     [Use  378  and  291.] 

18.  In  equivalent  triangles  the  bases  and  the  altitudes  upon  them  are 
reciprocally  proportional. 

19.  If  two  isosceles  triangles  have  the  legs  of 
one  equal  to  the  legs  of  the  other,  and  the  vertex- 
angle    of  the  one  the  supplement  of  the  vertex- 
angle  of  the  other,  the  tfiangles  are  equivalent. 

Given :  A  AB  C  and  A  CD,  etc. 

20.  Two  triangles  are  equivalent  if  they 
have  two  sides  of  one  equal  to  two  sides  of 
the  other  and  the  included  angles  supplemen- 
tary. 

Proof:  ZCAD  =  ^C'AD'  (?)  and  CA  =      L 
C'A  (V).     /,  the  rt.  A  are  =  (?).    Etc. 

21.  If  two  triangles  have  an  angle  of  one  the 
supplement  of  an  angle  of  the  other,  the  triangles 

are  to  each  other  as  the  products  of  the  sides  in-          \ 
eluding  these  angles.  C 

Given :  A  ABD  and  EEC,  A  at  B  supplementary. 

Proof :   Draw  DC,  use  A  BCD,  and  proceed  as  in  388. 

22.  The  area  of  a  triangle  is  equal  to  half 
the  perimeter  of  the  triangle  multiplied  by  the 
radius  of  the  inscribed  circle. 

Proof:  Draw OA,  etc.  AAOC  = 
&AOB  =  \AE  •  r  (?),  etc.     Add. 

23.  The  area  of  a  polygon  circumscribed  about  a  circle  is  equal  to 
half  the  product  of  the  perimeter  of  the  polygon  by  the  radius  of  the 
circle. 

24.  The  line  joining  the  midpoints  of  the  bases  of  a  trapezoid  bisects 
the  area  of  the  trapezoid. 

25.  Any  line  drawn  through  the  midpoint  of  a  diagonal  of  a  paral- 
lelogram, intersecting  two  sides,  bisects  the  area  of  the  parallelogram. 

26.  The  lines  joining  (in  order)  the  midpoints  of  the  sides  of  any 
quadrilateral  form  a  parallelogram  whose  area  is  half  the  area  of  the 
quadrilateral. 


BOOK   IV  199 


27.  If  any  point  within  a  parallelogram  is  joined  to  the  four  vertices, 
the  sum  of  one  pair  of  opposite  triangles  is  equivalent  to  the  sum  of  the 
other  pair;  that  is,  to  half  the  parallelogram. 

28.  Is  a  triangle  bisected  by    an  altitude?     By  the  bisector  of  an 
angle  ?    By  a  median  ?    By  the  perpendicular  bisector  of  a  side  ?    Give 
reasons. 

29.  If  the  three  medians  of  a  triangle  are 
drawn,  there  are  six  pairs  of  triangles  formed, 
one  of  each  pair  being  double  the  other. 

To  Prove:  A  AOB  =  2  A  A  OE-,  etc. 

**  E  ** 

30.  If  the  midpoints  of  two  sides  of  a  tri- 
angle are  joined  to  any  point  in  the  base,  the  quadrilateral  formed  is 
equivalent  to  half  the  original  triangle. 

31.  If  lines  are  drawn  from  the  midpoint 
of  one  leg  of  a  trapezoid  to  the  ends  of  the 
other  leg,  the  middle  triangle  thus  formed  is 
equivalent  to  half  the  trapezoid. 

Proof:  Draw  median  EF=m.     Then  EF 
is  h  to  the  bases  (?).     Denote  the  altitude  of 

the  trapezoid  by  h.     Then  EF  bisects  h  (?).     A  BFE  =  \  m  .  \  h  (?). 
A  AEF  =  I  m  -  %h  (?).     .'.&ABE  =  %  mh.     Consult  387. 

32.  The  area  of   a  trapezoid  is  equal  to  the  product  of  one  of  the 
non-parallel  sides,  by  the  perpendicular  upon  it  from  the  midpoint  of  the 
other. 

Proof :  Prove  that  A  ABE  =  half  the  trapezoid,  by  No.  31.     But  the 
&ABE  =  \ABx  the  ±  to  AB  from  E  (?). 

.-.  half  the  trapezoid  =  \  AB  x  this  _L  (Ax.  1).     Etc. 

33.  If  through  the  midpoint  of  one  of  the 
non-parallel    sides  of   a  trapezoid  a   line   is 
drawn  parallel  to  the  other  side,  the  parallelo- 
gram formed  is  equivalent  to  the  trapezoid. 

34.  If  two   equivalent  triangles  have  an 
angle  of  one  equal  to  an  angle  of  the  other, 

the  sides  including  these  angles  are  reciprocally  proportional. 

35.  The  sum  of  the  three  perpendiculars  drawn  to  the  three  sides 
of  an  equilateral  triangle  from  any  point  within  is  constant  (being  equal 
to  the  altitude  of  the  triangle). 

Proof:  Join  the  point  to  the  vertices.      Set  the  sum  of  the  areas  of 
the  three  inner  A  equal  to  the  area  of  the  whole  A.    Etc, 


200 


PLANE   GEOMETRY 


36.   In  the  figure  of  391,  prove  : 
(t)   Points  /,  C,  and  F  are  in  a  straight  line. 
(ft)  CE  and  BF  are  perpendicular. 
(fff)   A  G  and  BH  are  parallel. 


3 

TO 

n 

m             m 

m 

m 

On 

n              n 

n 

m 

n 

1 

- 

[See  Ex.  20.] 

37.  The  sum  of  the  squares  described  upon 
the  four  segments  of  two  perpendicular  chords  in 
a  circle  is  equivalent  to  the  square  described  upon 
the  diameter.    (Fig.  is  on  page  178.) 

38.  The  square  described  upon  the  sum  of  two 
lines  is  equivalent  to  the  sum  of  the   squares  de- 
scribed upon  the  two  lines,  plus  twice  the  rectangle 
of  these  lines. 

To  Prove  :  Square  A  E  =c=  m2  +  n2  +  2  mn. 

39.  The  square  described  upon  the  difference  of 
two  lines  is  equivalent  to  the  sum  of  the  squares  de- 
scribed upon  the  two  lines  minus  twice  the  rectangle 
of  these  lines. 

To  Prove  :  Square  AD  =  m*  +  n2  -  2  mn. 

40.  A  and  B  are  the  extremities  of  a  diameter 
of  a  circle  ;  C  and  D  are  the  points  of  intersection  of 
any  third  tangent  to  this  circle,  with  the  tangents 
at  A    and  B  respectively.     Prove  that  the  area  of 
ABDC  is  equal  to  £  AB  •  CD. 

41.  If  the  four  points  midway  between  the  center  and  vertices  of  a 
parallelogram  be  joined  in  order,  there  will  be  a  parallelogram  formed  ; 
it  will  be  similar  to  the  original  parallelogram  ;  its  perimeter  is  half  of 
the  perimeter  of  the  original  figure  ;  and  its  area  is  one  quarter  of  the 
area  of  the  original  figure. 

42.  If  two  equivalent  triangles  have  the  same  base  and  lie  on  opposite 
sides  of  it,  the  line  joining  their  vertices  is  bisected  by  the  base. 

43.  What  part  of  a  right  triangle  is  the  quadrilateral  which  is  cut 
from  the  triangle  by  a  line  joining  the  midpoints  of  the  legs? 

44.  Show  by  drawing  a  figure  that  the  square  on  half  a  line  is  one 
fourth  the  square  on  the  whole  line. 

45.  From  M,  a  vertex  of  parallelogram  LMNO,  a  line  MPX  is  drawn 
meeting  NO  at  P  and  LO  produced,  at  X.     LP  and  NX  are  also  drawn. 
Prove  triangles  LOP  and  XNP  are  equivalent. 


G 
C 

-n-l 
n      n 
n 

m-n 
m-n         m-n 
m-n 

m-n 
n 

[ 

n 

m 

D 
n 

BOOK  IV 


201 


FORMULAS 

397.   PROBLEM.   To  derive  a  formula  for  the  area  of  a  triangle  in 
terms  of  its  sides. 

Given :  A  ABC,  having   sides  A 


Required:  To  derive  a  for- 
mula for  its  area,  containing 
only  a,  6,  and  e.  c 


Solution :  Draw  altitude  AD. 
Now  CD  =  bpa=  <**+&-<?  (349), 

and        AD2  =  A(?-cJ?  (392). 


Hence,  A»= 
and          F= 


b  —  -  -  I  (by  factoring), 


2a 


_  (a  +  b  +  g)  (a  +  b  -  g)  (g  +  a  -  5)  (g  -  a  +  ft) 


,  _ 
= 


4a2 


Now,  area  of  A  =  -  a  -  h  = 


as 
i(  a 

•2"  VL 


4a2 


).  Q.E.F. 


To  simplify  this  formula,  let  us  call  a 
Then,  it  is  evident  that  a  +  b—  <?=2(s  —  <?); 

=  2(s  — a). 


202  PLANE   GEOMETRY 

Substituting,  in  the  formula  for  area, 


Area  of  A  =  £  V2  s  -  2  (s  —  c)  -  2  (*  -  6)  •  2  (*  —  a). 
That  is, 

Area  of  A  =  Vs(s  -a)(s-  6)  (s  -c). 


EXERCISE.   Find  the  area  of  a  triangle  whose  sides  are   17,  25,  28. 
Here,  a  =  17,  b  =  25,  c  =  28,  5  =  35,  s  -  a  =  18,  s  -  b  =  10,  s  -  c  =  7. 
Area  =  V35  •  18  •  10  •  7  =  V72  .  52 .  22  .  32  =  210. 


398.   PROBLEM.    To  derive  formulas  for  the  altitudes  of  a  triangle 
in,  terms  of  the  three  sides. 


Solution:  Area  =  J  aha  =  Vs  (s  —  a)  (s  —  6)  (s  — 


-  «)  (8  -  ft)  Q  -  C) 

la 


Similarly,  fe6  = 


399.   PROBLEM.   To   derive  the  formulas  for  the  altitude  and  the 
area  of  an  equilateral  triangle,  in  terms  of  its  side. 

Solution :    Let  each  side  =  #, 
and  altitude  =  h. 


Then,  A2=  «2-      =        (?). 


Also, 
Area=ibase.^  =  !a-?V3. 

.*.  Area  = 


Ex.  1.  Find  the  area  of  the  triangle  whose  sides  are  7,  10,  11. 

Ex.  2.  Find  the  area  of  the  triangle  whose  sides  are  8,  15,  17. 

Ex.  3.  Find  the  area  of  the  equilateral  triangle  whose  side  is  8. 

Ex.  4.  Find  the  side  of  the  equilateral  triangle  whose  area  is  121 V3- 

Ex.  5.  Find  the  area  of  the  equilateral  triangle  whose  altitude  is  10. 


BOOK  IV 


203 


400.   PROBLEM.   To  derive  the  formula  for  the  radius  of  the  circle 
inscribed  in  a  triangle,  in  terms  of  the  sides  of  the  triangle. 


Solution 


Adding, 
(Because, 

Hence,  r 


B 


Area  of  A  A  OB  —  \e  •  r\ 
area  of  A  AOC  —  ^b-r  l(?). 
area  of  A  BOC—  \a-r\ 

area  of  A  ABC  =  ^(a  +  b  +  c)r 
(a  +  b  +  c)  =  s.) 
area  of  A  ABC 


r  _- 

— 


Vs  (9  - 


-  b}  (s- 


401.  PROBLEM.  To  derive  the 
formula  for  the  radius  of  the  circle 
circumscribed  about  a  triangle,  in 
terms  of  the  sides  of  the  triangle. 

Solution : 
2B  -  ha  =  b  -  c  (?)  (337). 

,.*-L-. 


h  ^ 


a  •  b  -  c 


Ex.  1.  Fiud  the  radius  of  the  circle  inscribed  in,  and  the  radius  of  the 
circle  circumscribed  about,  the  triangle  whose  sides  are  17,  25,  28. 

Ex.  2.  Find  for  triangle  whose  sides  are  11,  14,  17,  the  radii  of  the 
inscribed  and  circumscribed  circles. 


204  PLANE  GEOMETRY 


ORIGINAL   EXERCISES  (NUMERICAL) 

1.  The  base  of  a  parallelogram  is  2  ft.  6  in.  and  its  altitude  is  1  ft. 
4  in.     Find  the  area.     Find  the  side  of  an  equivalent  square. 

2.  The  area  of  a  rectangle  is  540  sq.  m.  and  its  altitude  is  15  m. 
Find  its  base  and  diagonal. 

3.  The  base  of  a  rectangle  is  3  ft.  4  in.  and  its  diagonal  is  3  ft.  5  in. 
Find  its  area. 

4.  The  bases  of  a  trapezoid  are  2  ft.  1  in.,  and  3  ft.  4  in.,  and  the 
altitude  is  1  ft.  2  in.     Find  the  area. 

5.  The  area  of  a  trapezoid  is  736  sq.  in.  and  its  bases  are  3  ft.  and 
4  ft.  8  in.     Find  the  altitude. 

6.  The  area  of  a  certain  triangle  whose  base  is  40  rd.,  is  3.2  A.    Find 
the  area  of  a  similar  triangle  whose  base  is  10  rd.     Find  the  altitudes  of 
these  triangles. 

7.  The  base  of  a  certain  triangle  is  20  cm.     Find  the  base  of  a  simi- 
lar triangle  four  times  as  large ;   of  one  five  times  as  large ;   twice  as 
large;  half  as  large;  one  ninth  as  large. 

8.  The  altitude  of  a  certain  triangle  is  12  and  its  area  is  100.     Find 
the  altitude  of  a  similar  triangle  three  times  as  large.     Find  the  base  of 
a  similar  triangle  seven  times  as  large.     Find  the  altitude  and  base  of  a 
similar  triangle  one  third  as  large. 

9.  The  area  of  a  polygon  is  216  sq.  m.  and  its  shortest  side  is  8  m. 
Find  the  area  of  a  similar  polygon  whose  shortest  side  is  10  m.     Find  the 
shortest  sid.e  of  a  similar  polygon  four  times  as  large;  one  tenth  as  large. 

10.  If  the  longest  side  of  a  polygon  whose  area  is  567  is  14,  what  is 
the  area   of   a  similar  polygon  whose   longest  side  is   12  ?   of  another 
whose  longest  side  is  21  ? 

11.  Find  the  area  of  an  equilateral  triangle  whose  sides  are  each  6  in. 
Of  another  whose  sides  are  each  10 V  3  ft. 

12.  Find  the  area  of  an  equilateral  triangle  whose  altitude  is  4  in. ; 
of  another  whose  altitude  is  18  dm. 

13.  The  area  of  an  equilateral  triangle  is  64 \/3.    Find  its  side  and  its 
altitude. 

14.  The  area  of  an  equilateral  triangle  is  90  sq.  m.     Find  its  altitude. 

15.  Find  the  side  of  aiv  equilateral  triangle  whose  area  is  equal  to  a 
square  whose  side  is  15  ft. 


BOOK  IV  205 

16.  The  equal  sides  of  an  isosceles  triangle  are  each  17  in.  and  the 
base  is  16  in.     Find  the  area. 

17.  Find  the  area  of  an  isosceles  right  triangle  whose  hypotenuse  is 
2  ft.  6  in. 

18.  Find  the  area  of  a  square  whose  diagonal  is  20  m. 

19.  There  are  two  equilateral  triangles  whose  sides   are  33  and  56 
respectively.     Find  the  side  of  the  third,  equivalent  to  their  sum.     Find 
the  side  of  the  equilateral  triangle  equivalent  to  their  difference. 

20.  There  are  two  similar  polygons  two  of  whose  homologous  sides 
are  24  and  70.     Find  the  side  of  a  third  similar  polygon  equivalent  to 
their  sum  ;  the  side  of  a  similar  polygon  equivalent  to  their  difference. 

21.  What  is  the  area  of  the  right  triangle  whose  hypotenuse  is  29 
cm.  and  whose  short  leg  is  20  cm.  ? 

22.  The  base  of  a  triangle  is  three  times  the  base  of  an  equivalent 
triangle.     What  is  the  ratio  of  their  altitudes? 

23.  The  bases  of  a  trapezoid  are  56  ft.  and  44  ft.  and  the  non- paral- 
lel sides  are  each  10  ft.     Find  its  area.     Also  find  the  diagonal  of  an 
equivalent  square. 

24.  The  base  of  a  triangle  is  80  m.,  and  its  altitude  is  8  m.    Find  the 
area  of  the  triangle  cut  off  by  a  line  parallel  to  the  base  and  at  a  dis- 
tance of  3  m.  from  it.     Another,  cut  off  by  a  line  parallel  to  the  base 
and  6  m.  from  it. 

25.  The  bases  of  a  trapezoid  are  30  and  55,  S\\ 
and  its  altitude  is  10.     If  the  non-parallel  sides                   >^      jx\ 
are  produced  till  they  meet,  find  the  area  of  the             >"^~  \ 
less  triangle  formed.  /          6g      "          \ 

[The  A  are  similar.    .'.30  :  55= x :  x  +  10.  Etc.] 

26.  The  diagonals  of  a  rhombus  are  2  ft.  and  70  in.     Find  the  area; 
the  perimeter  ;  the  altitude. 

27.  The  altitude  (Ji)  of  a  triangle  is  increased  by  n  and  the  base  (&) 
is  diminished  by  x  so  the  area  remains  unchanged.     Find  x. 

28.  The  projections  of  the  legs  of  a  right  triangle  upon  the  hypote- 
nuse are  8  and  18.     Find  the  area  of  the  triangle. 

29.  In  triangle  ABC,  AB  is  5,  BC  is  8,  and  AB  is  produced  to  P, 
making  BP=Q.     BC  is  produced  (through  5)  to  Q  and  PQ  drawn  so  the 
triangle  BPQ  is  equivalent  to  triangle  ABC.      Find  the  length  of  BQ. 
[Use  388.] 


206  PLANE   GEOMETRY 

30.  The  angle  C  of  triangle  ABC  is  right ;  A  C  =  5;  EC  =  12.    BA 
is  produced  through  A,  to  D  making  AD  =  4;  CA  is  produced  through 
A,  to  E  so  triangle  AED  is  equivalent  to  triangle  ABC.     Find  ^4  #. 

31.  Find  the  area  of  a  square  inscribed  in  a  circle  whose  radius  is  6. 

32.  Find  the  side  of  an  equilateral  triangle  whose  area  is  25  V3". 

33.  Two  sides  of  a  triangle  are  12  and  18.      What  is  the  ratio  of  the 
two  triangles  formed  by  the  bisector  of  the  angle  between  these  sides? 

34.  The  perimeter  of  a  rectangle  is  28  m.  and  one  side  is  5  m.     Find 
the  area. 

35.  The  perimeter  of  a  polygon  is  5  ft.  and  the  radius  of  the  inscribed 
circle  is  5  in.     Find  the  area  of  the  polygon. 

In  the  following  triangles,  find  the  area,  the  three  altitudes,  radius  of 
inscribed  circle,  radius  of  circumscribed  circle : 

36.  a  =  13,  b  =  14,  c  =  15. 

37.  a  =  15,  b  =  41,  c  =  52. 

38.   20,  37,  51.  39.   25,  63,  74.  40.    140,  143,  157. 

41.  The  sides  of  a  triangle  are  15,  41,  52 ;  find  the  areas  of  the  two 
triangles  into  which  this  triangle  is  divided  by  the  bisector  of  the  largest 
angle. 

42.  Find  the  area  of  the  quadrilateral  A  BCD  if  AB  =  78  m.,  EC  - 
104m.,  CD  =  50  m.,  AD  =  120  m.,  and  AC  =  130  m. 

43.  One  diagonal  of  a  rhombus  is  ^  of  the  other  and  the  difference 
of  the  diagonals  is  14.     Find  the  area  and  perimeter  of  the  rhombus. 

44.  A  trapezoid  is  composed  of  a  rhombus  and  an  equilateral  triangle ; 
each  side  of  each  figure  is  16  inches.     Find  the  area  of  the  trapezoid. 

45.  Find  the  side  of  an  equilateral  triangle  equivalent  to  the  square 
whose  diagonal  is  15  V2. 

46.  Which  of  the  figures  in  No.  45  has  the  less  perimeter? 

47.  In  a  triangle  whose  base  is  20  and  whose  altitude  is  12,  a  line  is 
drawn  parallel  to  .the  base,  bisecting  the  area  of  the  triangle.    Find  the 
distance  from  the  .base  to  this  parallel. 

48.  Parallel  to  the  base  of  a  triangle  whose  base  is  30  and  altitude  is 
18  are  drawn  two  lines  dividing  the  area  of  the  triangle  into  three  equal 
parts.     Find  their  distances  from  the  vertex. 

49.  Around  a  rectangular  lawn  30  yards  x  20  yards  is  a  drive  16  feet 
wide.    How  many  square  yards  are  there  in  the  drive  ? 


4A2     Pan 


BOOK    IV 
CONSTRUCTIONS 


207 


402.  PROBLEM.    To  construct  a  square  equivalent  to  the  sum  of  two 
squares. 


A  B          C  D         E  F 

Given  :  (?).  Required  :  (?).  Construction  :  Construct  a 
rt.  Z  E,  whose  sides  are  EX  and  ET.  On  EX  take  EF  =  AB, 
and  on  ET  take  EG=  CD.  Draw  FG.  On  FG  construct  square  T. 

Statement:    T^B+S.  Q.B.F. 

Proof  :  ~GF2  =  EF*  +  2?(?  (?).  But  GF2  =  T  ;  EF2  =  £  ; 
EG2  =  5  (?)  (372).  Hence,  r  =c=  u  +s  (Ax.  6).  Q.E.D. 


403.  PROBLEM.   To  construct  a  square  equivalent  to  the  sum  of 
several  squares. 

Given  :  Squares  whose  sides  are 
a,  b,  <?,  d. 

Required  :  To  construct  a  square 


Construction  :  Construct  a  rt.  Z. 
whose  sides  are  equal  to  a  and  b. 
Draw  hypotenuse  BC.  At  B  erect 
a  J_  =  G  and  draw  hypotenuse,  DC. 
At  D  erect  a  J_  =  c?,  etc. 

Statement:    The   square    con-    ° 
structed  on  EC  is  =o  to  the  sum  of 
the  several  given  squares. 

Proof:     EC2  =DC2  +  d*  =  (BC2  + 


+  d? 


Q.E.F. 


.  Q.E.D. 


208 


PLANE   GEOMETRY 


404.   PROBLEM.    To  construct  a  square  equivalent  to  the  difference 
of  two  given  squares. 


R 


B     C 


Given:  (?).     Required:  (?). 

Construction :  At  one  end  of  indefinite  line,  EX,  erect  EG 
X  to  EX  and  =  CD  (a  side  of  the  less  square,  s).  Using  G  as 
center  and  AB  as  radius,  describe  arc  intersecting  EX  at  F. 

Draw  GF.     On  EF  construct  square  T. 

Statement:  T^R  —  S.     Q.E.F.  Proof:  (?). 

405.  PROBLEM.   To   construct  a   polygon   similar  to  two  given 
similar  polygons  and  equivalent  to  their  sum. 

Construction:  Like  402.     Proof:   (393). 

406.  PROBLEM.   To  construct  a   polygon   similar  to  two  given 
similar  polygons  and  equivalent  to  their  difference. 

Construction:  Like  404.     Proof:   (394). 

407.  PROBLEM.   To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 


A  B        A1  B1  E'    X  B' 

Given:  (?).     Required:   (?).       Construction:   Construct  a 


BOOK  IV 


209 


mean  proportional  between  the  base,  AB,  and  the  altitude, 
BE\  on  this  mean  proportional,  B'G,  construct  a  square,  M. 

Statement:    Square  M  =c=  parallelogram  P.  Q.E.F. 

Proof  :    AB  :  BrG  =  B'G  :  BE  (Const.). 

.-.  WG2  =  AB-BE  (?).      But  We?  =  M  (?). 

Aiid^£.£#  =  P(?)(371).     Hence,  Jf=c=P( Ax.  6).  Q.E.D. 

408.  PROBLEM.   To  construct  a  square  equivalent  to  a  given  tri- 
angle. 

Construct  a  mean  proportional  between  half  the  base  and 
the  altitude,  and  proceed  as  in  407. 

409.  PROBLEM.   To  construct  a  triangle  equivalent  to  a  given  poly- 
gon. 


A  B G          I         A  G          H 

Given :  Polygon  AD.     Required  :    To  construct  a  A  =0=  AD. 

Construction  :  Draw  a  diagonal,  BD,  connecting  any  vertex 
(JS)  to  the  next  but  one  (Z>).  From  the  vertex  between 
these  (C),  draw  CG  II  to  BD,  meeting  AB  prolonged,  at  G. 
Draw  DG.  Repeat  (2d  figure)  by  drawing  EG,  then  DH  II  to 
EG,  and  EH.  Repeat  again  by  drawing  AE,  FI II  to  AE,  and  El. 

Statement:  A  IEH o=  polygon  ABCDEF.  Q.E.F. 

Proof  :   In  1st  fig.,  A  BGD  =e=  A  BCD  (381 ;  BD  is  the  base). 

Add          polygon  ABDEF  =  polygon  ABDEF. 

.'.polygon  AGDEF  =c=  polygon  ABCDEF  (Ax.  2). 

Likewise,  AHEF=o=  AGDEF -,  and  A  lEH^AHEF.   (Explain.) 

.'.  A  IEH  =0=  polygon  ABCDEF  (Ax.  1).  Q.E.D. 

410.  PROBLEM.  To  construct  a  square  equivalent  to  a  given  poly- 
gon. Tllse  409  and  408.] 


210 


PLANE   GEOMETRY 


411.   PROBLEM.   To    construct  a  parallelogram    (or  a  rectangle) 
equivalent  to  a  given  square,  and  having : 

I.  The  sum  of  its  base  and  altitude  equal  to  a  given  line. 
II.  The  difference  of  its  base  and  altitude  equal  to  a  given  line. 


c1 


E 

:       F 

>::  ••••...      \ 

V 

..     \ 

\ 

s 

\     c 

;•                                D' 

A  B     C  G 

I.  Given  :  Square  S  and  line  CD. 

Required :  To  construct  a  O  o  S,  whose  base  +  altitude 
shall  =  CD. 

Construction :  On  CD  as  a  diameter  describe  a  semicircle. 
At  C  erect  CE  _L  to  CD  and  =  AB.  Through  E  draw  EF  II  to 
CD,  meeting  the  circumference  at  F.  Draw  FG  _L  to  CD. 
Take  G'D'=  GD  and  draw  XY  II  to  GrDr  at  the  distance  from 
it  =  CG.  On  XY  take  HI  =  GD.  Draw  HGf  and  ID' . 

Statement:  O  G'D'IH  ^  S  and  base  +  alt.  =  CD.        Q.E.F. 

Proof :  G'D'lHis  a  a  (?)  (135)._  GDXCG=F(?  (?)  (341). 

But  GD  x  C£=area  G'D'IH  (?).    FG2=^c2  =  area  S.  (Explain.) 

/.  /U  G'DrIHoS(Ax.  6).    Also  (?'!>'  +(?'(/  =  CD  (?).     Q.E.D. 


X      H 


Cr 


X..  H r 

!o^ 

:  E' 


II.  Given :  Square  5  and  line  CD. 

Required  :  To  construct  a  O  =^  s ;  base  —  altitude  =  C-D. 


BOOK  IV 


211 


Construction  :  On  CD  as  diameter,  describe  a  O,  o.  At  C 
erect  CE  A-  to  CD  and  —  AB.  Draw  EFOG  meeting  O  at  F  and 
G.  Take  E'Gf  =  EG  and  draw  XY  II  to  E'G'  at  a  distance 
from  it  =  EF.  On  XY  take  HI  =  .EG.  Draw  HE'  and  1C?'. 

Statement:  £7  E'G'IH^  s  and  base  —  alt.  =  CD.          Q.E.F. 

Proof:  M7  is  tangent  to  Oo(?).    .'.EG  •  EF=EC2(?)  (333). 

EG  •  EF=  area  O  E'G'IH  (?),  and  #C2  =  JJ?  =  area  S  (?). 

FG  =  CD(?.  Q.E.D. 


412.   PROBLEM.   To  find  two  lines  whose  product  is  given  : 
I.  If  their  sum  is  also  given.  e  game  ag 

II.  If  their  difference  is  also  given, 


1  [Th 
,  j 


413.   PROBLEM.  To  construct  a  square  having  a  given  ratio  to  a 
given  square. 


D 
""  ••••-.. 


R 

.--'           S      \  *'*'•% 
a  ••'''           \X  \ 
\     \ 
/    E./  .*F\ 

S 

*  !••"*'          m             \        n         '-I         v 

X 

Q 

A                              B                    C 

i7?                                                  _/ 

Given  :    Square  E,  and  lines  m  and  n. 

Required  :    To  construct  a  square  such  that, 

The  square  R  :  the  unknown  square  =  m  :  n. 

Construction  :  On  an  indefinite  line  A  Y  take  AB  =  m,  and 
BC  =n.  On  AC  as  diameter  describe  a  semicircle.  At  B 
erect  #D_L  to  J.C,  meeting  arc  at  D.  Draw  AD  and  DC.  On 
^1Z>  take  DE=a,  and  draw  .£F  H  to  AC,  meeting  DC  at  F. 
Using  DF  =#,  as  a  side,  construct  square  S. 

Statement :    R  :  S  =  m  :  n.  Q.E.F. 

Proof  :  Z.  ADC  is  a  rt.Z(?).    .-.  ii?  :  DC2  =  m  :  n  (?)  (395). 


-:=^?);  •••p=^<^)- 


(Ax.  6). 


Q.E.D. 


PLANE   GEOMETRY 


414.   PROBLEM.  To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  given  ratio  to  it. 


Given  :    (?).     Required  :    (?).   Construction  and  Statement 
are  the  same  as  in  413. 

Proof  :   ^  ADCis  a  rt.  Z  (?).     .-.Al?:~J)C2  =  m:n  (?)(395). 

«=^(?);...^  =  ^2  =  ^.  (Explain.)    *  =  ^  (?)(390). 
x     DC^  &     w*      n    ^  S     x*  ^ 

.'.  E  :  S  —  m  :  n  (Ax.  1).  Q.E.D. 

415.    PROBLEM.  To  construct  a  polygon  similar  to  one  given  poly- 
gon and  equivalent  to  another. 


A—      — B  a 3 

Given  :    Polygons  B  and  s.    Required :    (?). 

Construction  :  Construct  squares  Rf=o=R,  and  Sr=o=S  (by  410). 
Find  a  fourth  proportional  to  a,  b,  and  AB.  This  is  CD, 
Upon  CD,  homologous  to  AB,  construct  T  similar  to  R. 

Statement:    r=o&  Q.E.F. 


CD' 


(.  |  =^5  (Const.).  /•-  =  - 
b      CD  b*      CD 


.   Now,  a2  = 


.-.5  =  5    (Ax.  6).     .\T^8    (Ax.  3). 


='s'=o=/S.  (Explain.) 

Q.E.D. 


BOOK  IV  213 

ORIGINAL   CONSTRUCTIONS 

1.  To  construct  a  square  equivalent  to  a  given  right  triangle. 

2.  To  construct  a  right  triangle  equivalent  to  a  given  square. 

3.  To  construct  a  right  triangle  equivalent  to  a  given  parallelogram. 

4.  To  construct  a  square  equivalent  to  the  sum  of  two  given  right 
triangles. 

5.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
right  triangles. 

6.  To  construct  a  square  equivalent  to  the  sum  of  two  given  paral- 
lelograms. 

7.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
parallelograms. 

8.  To  construct  a  square  equivalent  to  the  sum  of  several  given  right 
triangles. 

9.  To  construct  a  square  equivalent  to  the  sum  of  several  given  paral- 
lelograms. 

10.  To  construct   a  square  equivalent  to  the  sum  of  several  given 
triangles. 

11.  To  construct  a  square  equivalent  to  the  sum  of  several  given 
polygons. 

12.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
polygons. 

13.  To  construct  a  square  equivalent  to  three  times  a  given  square. 
To  construct  a  square  equivalent  to  seven  times  a  given  square. 

14.  To   construct  a  right  triangle  equivalent  to  the  sum  of  several 
given  triangles. 

15.  To  construct  a  right  triangle  equivalent  to  the  difference  of  any 
two  given  triangles;  of  any  two  given  parallelograms. 

16.  To  construct  a  square  equivalent  to  a  given  trapezoid  ;  equiva- 
lent to  a  given  trapezium. 

17.  To  construct  a  square  equivalent  to  a  given  hexagon. 

18.  To  construct    a   rectangle   equivalent  to   a  given  triangle,  hav- 
ing given  its  perimeter. 

19.  To  construct  an   isosceles  right  triangle  equivalent  to  a  given 
triangle. 

20.  To  construct  a  square  equivalent  to  a  given  rhombus. 

21.  To  construct  a  rectangle  equivalent  to  a  given   trapezium,  and 
having  its  perimeter  given, 


214  PLANE   GEOMETRY 

22.  To  find  a  line  whose  length  shall  be  V2  units.     [See  402.] 

23.  To  find  a  line  whose  length  shall  be  V3  units. 

24.  To  find  a  line  whose  length  shall  be  v/II  units. 

25.  To  find  a  line  whose  length  shall  be  Vf  units. 

26.  To  find  a  line  whose  length  shall  be  VlO  units. 

27.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

28.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

29.  To  construct  a  polygon  which  shall  be  f  of  a  given  polygon,  and 
similar  to  it. 

30.  To  construct  a  square  which  shall  have  to  a  given  square  the 
ratio  V3  :  4.     If  the  given  ratio  is  4  :  VF. 

31.  To  draw  through  a  given  point,  within  a  parallelogram,  a  line 
which  shall  bisect  the  parallelogram. 

32.  To  construct  a  rectangle  equivalent  to  a  given  trapezoid,  and 
having  given  the  difference  of  its  base  and  altitude. 

33.  To  construct  a  triangle  similar  to  two  given  similar  triangles  and 
equivalent  to  their  sum. 

34.  To  construct  a  triangle  similar  to  a  given  triangle  and  equivalent 
to  a  given  square.     [See  415.] 

35.  To  construct  a  triangle  similar  to  a  given  triangle  and  equivalent 
to  a  given  parallelogram. 

36.  To  construct  a  square  having  twice  the  area  of  a  given  square. 
[Two  methods.] 

37.  To  construct  a  square  having  3|  times  the  area  of  a  given  square. 

38.  To  construct  an  isosceles  triangle  equivalent  to  a  given  triangle 
and  upon  the  same  base. 

39.  To  construct  a  triangle  equivalent  to   a  given  triangle,  having 
the  same  base,  and  also  having  a  given  angle  adjoining  this  base. 

40.  To  construct   a  parallelogram  equivalent    to  a  given  parallelo- 
gram, having  the  same  base   and  also  having  a  given  angle  adjoining 
the  base. 

41.  To  draw  a  line  that  shall  be  perpendicular  to  the  bases  of  a 
parallelogram  and  that  shall  bisect  the  parallelogram. 

42.  To  construct  an  equilateral   triangle  equivalent   to  a  given  tri- 
angle.    [See  415.] 

43.  To  trisect   (divide  into  three  equivalent  parts)  the  area  of  a 
triangle,  by  lines  drawn  from  one  vertex. 


BOOK  IV  215 

44.  To  construct  a  square  equivalent  to  f  of  a  given  pentagon. 

45.  To    construct    an    isosceles    trapezoid    equivalent    to    a    given 
trapezoid. 

46.  To  construct  an  equilateral  triangle  equivalent  to  the  sum  of  two 
given  equilateral  triangles. 

47.  To  construct  an  equilateral  triangle  equivalent  to  the  difference 
of  two  given  equilateral  triangles. 

48.  To  construct    upon   a   given   base    a   rectangle   that    shall   be 
equivalent  to  a  given  rectangle. 

Analysis :  Let  us  call  the  unknown  altitude  x. 
Then  b-h  =  b'-x  (V).     Hence,  b':b  =  h:x(t). 


A 

b 


That  is,  the  unknown  altitude  is  a  fourth  pro- 
portional to  the  given  base,  the  base  of  the  given  rec- 
tangle, and  the  altitude  of  the  given  rectangle. 

Construction:  Find  a  fourth  proportional,  #,  to  &',  b  and  h.     Con- 
struct a  rectangle  having  base  =  b'  and  alt.  =  x. 

Statement:  This  rectangle,  B^A.  I n 

Proof:  b':b  =  h:x  (Const.).  .-.  Vx  =  bh  (?).   But    L I 

b'x  =  the  area  of  B  (?).  Etc. 

49.  To  construct  a  rectangle  that  shall  have  a  given  altitude  and  be 
equivalent  to  a  given  rectangle. 

50.  To  construct  a  triangle  upon  a  given  base  that  shall  be  equiva- 
lent to  a  given  triangle. 

51.  To  construct  a  triangle  that  shall  have  a  given  altitude  and  be 
equivalent  to  a  given  triangle. 

52.  To  construct  a  rectangle  that  shall  have  a  given  base,  and  shall 
be  equivalent  to  a  given  triangle. 

53.  To  construct  a  triangle  that  shall  have  a   given  base,  and  be 
equivalent  to  a  given  rectangle. 

54.  To  construct  a  triangle  that  shall  have  a  given  base  and  be 
equivalent  to  a  given  polygon. 

55.  Construct  the  problems  49,  50,  51,  53,  54  if  the  first  noun  in  each 
problem  is  the  word  "  parallelogram." 

56.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equivalent 
to  a  given  triangle. 

57.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equivalent 
to  a  given  square. 


216  PLANE   GEOMETRY 

58.  To  construct  a  triangle  which  shall  have  a  given  base,  a  given 
adjoining  angle,  and  be  equivalent  to  a  given  triangle.     Another,  equiva- 
lent to  a  given  square.     Another,  equivalent  to  a  given  polygon. 

59.  To  construct  a  parallelogram  which  shall  have  a  given  base,  a 
given   adjoining  angle,   and   be  equivalent  to   a  given   parallelogram. 
Another,  equivalent  to  a  given  triangle;  to  a  given  polygon. 

60.  To  construct  a  line,  DE,  from  D  in  AB  of  triangle  ABC,  so  that 
DE  bisects  the  triangle.  A 

Analysis:  After  DE  is  drawn,  &ABC=2&ADE 

Hence,  AB  •  A  C  =  2  (AD  .  A  E)  (Ax.  6). 

/.  2  AD :  AB  =  A  C :  x  (?).     Thus  x,  (that  is,  AE)  is 
a  fourth  proportional  to  three  given  lines. 

61.  To  draw  a  line  meeting  two  sides  of  a  triangle  and  forming  an 
isosceles  triangle  equivalent  to  the  given  triangle. 

Analysis:  Suppose  AX  &  leg  of  isosceles  A. 
.:AABC:&AXX'=AB.AC:AX.AX'. 
But  the  A  are  equivalent  and  AX  =  AX'    (Hyp.). 
Hence,  AB  -  AC  —  A  X2.   :.AX  is  a  mean  propor-     B~ 
tional  between  AB  and  A  C. 

62.  To  draw  a  line  parallel  to  the  base  of  a  triangle  which  shall 
bisect  the  triangle.     [See  389  and  use  414.] 

63.  To  draw   a  line   meeting  two  sides  of  a  triangle  forming  an 
isosceles  triangle  equivalent  to  half  the  given  triangle. 

64.  To  draw  a  line  parallel  to  the  base  of  a  triangle  forming  a  tri- 
angle equivalent  to  one  third  the  original  triangle. 

65.  To  draw  a  line  parallel  to  the  base  of  a  O 
trapezoid  so  that  the  area  is  bisected.                                              /\ 

Analysis:   A  OXX'  -  i   (A  OAD  +  A  OBC)  /     \ 

and  is  similar  to  A  OBC.  &Z. \C 

Construction :  [Use  408,  402,  415.] 

66.  To  construct  two  lines  parallel  to  the  base 

of  a  triangle,  that  shall  trisect  the  area  of  the    A 
triangle. 

67.  To  construct  a  triangle  having  given  its  angles  and  its  area. 
Analysis :  The  required  A  is  similar  to  any  A  containing  the  given 

A.     The  given  area  may  be  a  square.     This  reduces  the  problem  to  415. 

68.  To  find  two  straight  lines  in  the  ratio  of  two  given  polygons. 


BOOK  V 

REGULAR   POLYGONS.     CIRCLES 

416.  A  regular  polygon  is  a  polygon  which  is  equilateral 
and  equiangular. 

417.  THEOREM.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

Given  :    AG,  an  equilateral 
inscribed  potygon. 

To  Prove  :    AG  is  regular. 

Proof  :    /.  A  is  measured  by 
I  arc  BGL  (?). 

Also  Z  B  is    measured    by 
\  arc  CHA  (?),  etc. 

Subtended  arcs,  AB,  BC,  CD, 
etc.,  are  all  =  (?). 

Hence,  arcs  BGL,  CHA,  DIB, 
etc.,  are  all  =  (Ax.  2). 

/.  Z.A  =  /.B  =  ZO=  etc.    (?). 

That  is,  the  polygon  is  equiangular. 

Therefore,  the  polygon  is  regular  (?)  (416).  Q.E.D. 

418.  THEOREM.  If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  arcs,  and  the  chords  of  these  arcs  be  drawn,  they  will 
form  an  inscribed  regular  polygon. 

Proof:    Chords  AB,  BC,  CD,  etc.  are  all  =  (?). 

Therefore,  the  polygon  is  regular  (?).  Q.E.D. 


Ex.  1.  What  is  the  usual  name  of  a  regular  3-gon?   of   a  regular 
4-gon  ?     Is  an  equiangular  inscribed  polygon  necessarily  regular  ? 

Ex.  2.   In  the  figure  of  417,  how  many  degrees  are  there  in  each  of  the 
arcs,  AB,  BC,  etc.?     How  many  degrees  are  there  in  each  angle? 

217 


218 


PLANE   GEOMETRY 


419.  THEOREM.   If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  parts,  and  tangents  be  drawn,  at  the  several  points  of 
division,  they  will  form  a  circumscribed  regular  polygon. 

Given  :  (?).  To  Prove :  (?). 

Proof  :    Draw    chords  AB, 
BC,  CD,  etc. 

In   A  ABH,  BCI,   CDJ,  etc.,    G 
AB  =  BC  =  CD  =  etc.  (?). 

Z  HAB  =  Z  HBA  =  Zl£C  = 
Z.ICB  =  Z  JCD  =  etc.  (?). 

.*.  these  A  are  isosceles  (?) 
(120),  and  =  (?). 
/.  ZH=  Z/=  Z  J=etc.  (?). 
That  is,  polygon  GJ  is  equi- 
angular. 

Also,  AH  =  HB  =  BI  =  1C 
=  CJ  =etc.  (?),  and  HI  =  u  =  JK  =  etc.  (?)  (Ax.  3). 

That  is,  polygon  GJis  equilateral ;  /.  it  is  regular  (?).  Q.E.D. 

420.  THEOREM.     If    the    cir- 
cumference of  a  circle  be  divided 
into  any   number  of  equal  parts 
and  tangents  be  drawn  at  their 
midpoints,  they  will  form  a  cir- 
cumscribed regular  polygon. 

Given:  (?).  To  Prove:  (?). 

Proof :    Arcs   AB,  BC,  CD, 
etc.  are  all   =  (?). 

Also,  arcs  AI,  IB,  BK,  KG,   R 
CM,  etc.  are  all  =  (?)  (Ax.  3). 

.-.  arcs  IK,  KM,  MO,  etc.  are 
all  =  (?)  (Ax.  3). 

Therefore,  the  polygon  is  regular  (?)  (419).  Q.E.D. 

421.  THEOREM    If  chords  be  drawn  joining  the  alternate  vertices 
of  an  inscribed  regular  polygon  (having  an  even  number  of  sides), 
another  inscribed  regular  polygon  will  be  formed.     (See  417.) 


BOOK  V 


219 


422.  THEOREM.    If  the  ver- 
tices   of   an   inscribed    regular 
polygon  be  joined  to   the  mid- 
points of  the  arcs  subtended  by 
the  sides,  another  inscribed  regu- 
lar polygon  will  be  formed  (hav- 
ing double  the  number  of  sides). 

423.  THEOREM.    If   tangents 
be  drawn  at  the  midpoints  of  the 
arcs  between  adjacent  points  of 
contact  of  the  sides  of  a  circum- 
scribed regular  polygon,  another  circumscribed  regular  polygon  will 
be  formed  having  double  the  number  of  sides  (?). 

424.  THEOREM.    The  perimeter  of  an  inscribed  regular  polygon  is 
less  than  the  perimeter  of  an  inscribed  regular  polygon  having  twice 
as  many  sides,  and  the  perimeter  of  a  circumscribed  regular  polygon 
is  greater  than  the  perimeter  of  a  circumscribed  regular   polygon 
having  twice  as  many  sides.     (Ax.  12.) 

425.  THEOREM.    Two    regular 
polygons  having  the  same  number 
of  sides  are  similar. 

Given  :  Regular  w-gons  AD 
and  A'Dr. 

To  Prove  :  They  are  simi- 
lar. 

rc-2^  180° 


Proof:  /.A 


Z.A' 


(n-2)  180° 


n 


(?).      .-.  Z  A  =  Z  A'    (?). 


Similarly,  Z  B  =  /.  Bf,  Z.  C=  Z  c',  etc. 
That  is,  these  polygons  are  mutually  equiangular. 
Also,  AB  =  BC  =  CD=  etc.;  AfBf  =  BfCf  =  C'D'  = 
/.  AB  :  A'B'  =  BC  :  b'c'  =  CD  :  C'D'  =  etc.  (Ax.  3). 
That  is,  the  homologous  sides  are  proportional. 
Therefore,  the  polygons  are  similar  (?). 


(?). 


.E.D. 


220 


PLANE    GEOMETRY 


426.  THEOREM.  A  circle  can  be  circumscribed  about,  and  a  circle  can 
be  inscribed  in,  any  regular  polygon. 

Given :  Regular  polygon 
ABCDEF. 

To  Prove :  I.  A  circle  can 
be  circumscribed  about  the 
polygon. 

II.  A  circle  can  be  in- 
scribed in  the  polygon. 

Proof:    I.     Through  three 
consecutive    vertices,    A,    B, 
and  C,  describe  a  circumfer- 
ence, whose  center  is  O.    Draw  radii  OA,  OB,  OC,  and  draw 
line  OD. 

In  A  AOB  and  COD,  AB  =  CD  (?)  ;  BO  =  CO  (?). 

Now  ZABC  =  ^BCD  (?)  (416).     ZotfC  =  Z  OCB  (?). 
Subtracting,  Z  ABO  =  Z.  OCD  (Ax.  2). 

/.A AOB  =  A  COD  (?).      .'.AO  =  OD  (?). 

Hence,  the  arc  passes  through  D,  and  in  like  manner  it 
may  be  proved  that  it  passes  through  E  and  F. 

That  is,  a  circle  can  be  circumscribed  about  the  polygon. 

II.    AB,  BC,  CD,   DE,  etc.    are  =  chords  (?)  (416). 

Therefore  they  are  equally  distant  from  the  center  (?). 

That  is,  a  circle  described,  using  O  as  a  center  and  OM  as  a 
radius,  will  touch  every  side  of  the  polygon. 

Hence  a  circle  can  be  inscribed.     (Def.  234.)  Q.E.D. 

427.  The  radius  of  a  regular  polygon  is  the  radius  of  the 
circumscribed  circle.     The  radius  of  the  inscribed  circle  is 
called  the  apothem.     The  center  of  a  regular  polygon  is  the 
common  center  of  the  circumscribed  and  inscribed  circles. 

428.  The  central  angle  of  a  regular  polygon  is  the  angle 
included  between  two  radii  drawn  to  the  ends  of  a  side. 

429.  THEOREM.   Each  central  angle  of  a  regular  w-gon  =  ^-^-  (?)• 


BOOK  V 


221 


430.   THEOREM.   Each  exterior  angle  of  a  regular  n-gon 


_36oc 


431.  THEOREM.  The  radius  drawn  to  any  vertex  of  a  regular  poly- 
gon bisects  the  angle  at  the  vertex.     (See  80.) 

432.  THEOREM.  The  central  angles  of  regular  polygons  having  the 
same  number  of  sides  are  equal.    (See  429.) 

433.  THEOREM.   If  radii  be  drawn  to  all  the  vertices  of  a  circum- 
scribed regular  polygon,  and  chords  be  drawn  connecting  the  points  of 
intersection  of  these  lines  with  the  circumference,  an  inscribed  regular 
polygon  of  the  same  number  of  sides  will  be  formed  and  the  sides  of  the 
two  polygons  Will  be  respectively  parallel. 

Given:  (?).    To  Prove:  (?). 

Proof :  Central  A  at 
O  are  all  =  (?). 

.-.the  intercepted  arcs 
are  all  =  (?). 

/.the  chords  A'B', 
B'C',  etc.  are  all=  (?). 

/.  the  inscribed  poly- 
gon is  regular  (?). 

Also,  A  AOB,  AfOBf, 
etc.  are  isosceles  (?). 

If  a  line  OX  be  drawn 
from  O,  bisecting  /.  AOB, 
it  is  J_  to  AB  and  to  A'B'  (?)  (57). 

/.  AB  is  ||  to  A'B'  (?).  Q.E.D. 


Ex.  1.  How  many  degrees  are  there  in  the  angle,  in  the  central  angle, 
and  in  the  exterior  angle  of  a  regular  hexagon?  a  regular  decagon? 
a  regular  15-gon  ? 

Ex.  2.   In  the  figure  of  433,  prove, 

(a)  Triangle  A  OB  similar  to  triangle  A' OB'. 

(b)  Triangle  XOB  similar  to  triangle  X'OB'. 


222 


PLANE   GEOMETRY 


434.  THEOREM.  The  perimeters  of  two  regular  polygons  having 
the  same  number  of  sides  are  to  each  other  as  their  radii  and  also  as 
their  apothems. 


B1 


Given  :  Regular  w-gons,  EG  whose  perimeter  is  P,  radius  B, 
apothem  r\  and  E'C'  whose  perimeter  is  P',  radius  u', 
apothem  rr. 

To  Prove  :  P  :  P'  =  E  :  Ef  =  r  :  rf. 

Proof:  Draw  radii  OB  and  orBf.  In  A  AOB  and  A'O'B', 

;  AO  =  BO  and  ^L'o'  =  J3' 


Hence,          ^  (Ax.  3). 
A'O1      B'0f 

.'.  A  AOB  is  similar  to  A  ArOrBr  (?)  (317). 

A  ~f>  "D  M 

•*.     f   f  =  —  =  —  (?).     Also,  the  polygons  are  similar  (?). 
A  B        R        T 


Q.E.D. 


P'      A'B'  p' 

435.  THEOREM.  The  areas  of  two  regular  polygons  having  the 
same  number  of  sides  are  to  each  other  as  the  squares  of  their 
radii  and  also  as  the  squares  of  their  apothems. 

Proof  :  If  K  and  K1  denote  their  areas,  we  have  : 


A 


(?)(390).          But 


AB 
A7^' 


'-          (Explain.) 


Q.E.D. 


BOOK  V 


223 


436.   THEOREM.   The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  the  perimeter  by  the  apothem. 

Given:  (?). 

To  Prove:  (?). 

Proof:  Draw  radii  to  all  the 
vertices,  forming  several  isos- 
celes triangles. 

Area  of  A  AOB  =  %AB  -r 

Area  of  A  BOC  =  ^BC 

Area  of  A  COD  =     CD 


etc.,  etc. 


Area  of  poly  gon=%(AB 

or,  area  =  J  P-r  (Ax.  6). 


A  B. 

CD  -f-  etc.)  >r  (?), 


Q.E.D. 


437.  THEOREM.  If  the  number  of  sides  of  an  inscribed  regular  poly- 
gon be  increased  indefinitely,  the  apothem  will  approach  the  radius  as 
a  limit.  E^-* ^^D 

Given :  -ZV-gon  FC  inscribed  in 
O  0 ;   apothem  =  r  ;  radius  =  R. 

To  Prove:  That  as  the 
number  of  sides  is  indefi- 
nitely increased,  r  approaches 
B  as  a  limit. 

Proof :  In  the  A  AOK, 


or  R  — 

Now  as  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  AB  will  be  indefinitely  decreased. 

Hence,  ^  AB,  or  AK,  will  approach  zero  as  a  limit. 

.'.  R  —  r  will  approach  zero  (because  R  —  r  <  AK). 

That  is,  r  will  approach  R  as  a  limit  (240).  Q.E.D. 

NOTE.    It  is  evident  that  if   the  difference   between   two  variables 
approaches  zero,  either 

(1)  one  is  approaching  the  other  as  a  limit ;  or 

(2)  both  are  approaching  some  third  quantity  as  their  limit. 


224  PLANE   GEOMETRY 

Ex.  1.  The  apothem  of  a  regular  polygon  perpendicular  to  a  side 
bisects  the  side  and  the  central  angle  of  the  polygon. 

Ex.  2.  Are  all  pairs  of  squares  similar?  Why?  Are  all  pairs  of 
rhombuses  similar  ?  Why  ?  Is  a  rhombus  a  regular  polygon  ?  Is  a 
rectangle  a  regular  polygon  ? 

Ex.    3.    What  is  the  area  of  a  square  whose  perimeter  is  40  inches  ? 

Ex.  4.  What  is  the  area  of  a  regular  hexagon  whose  side  is  8  and 
apothem  is  4  V3  ? 

Ex.    5.   What   is  the   area   of  a  regular   dodecagon    whose  side    is 

10  ^2  -  V3  and  whose  apothem  is  5  \2  +  \/3  ? 

Ex.  6.  Prove  that  the  lines  joining  the  midpoints  of  the  sides  of  a 
regular  polygon,  in  order,  form  a  regular  polygon  of  the  same  number 
of  sides. 

Ex.  7.  Prove  that  a  polygon  is  regular  if  the  inscribed  and  circum- 
scribed circles  are  concentric. 

Ex.  8.  Draw  a  figure  showing  an  inscribed  equiangular  polygon  that 
is  not  regular. 

Ex.  9.  Draw  a  figure  showing  a  circumscribed  equilateral  polygon 
that  is  not  regular. 

Ex.  10.  Prove  that  the  circumference  of  a  circle  is  greater  than 
the  perimeter  of  any  inscribed  polygon. 


438.  The  theorems  of  439  and  440  are  considered  so  evident,  and 
rigorous  proofs  are  so  difficult  for  young  students  to  comprehend  (like 
the  demonstrations  for  many  fundamental  principles  in  mathematics) 
that  it  is  advisable  to  omit  the  profound  demonstrations  and  insert  only 
simple  explanations. 

439.  THEOREM.   The  circumference  of  a  circle  is  less  than  the  per- 
imeter of  any  circumscribed  polygon. 

By  drawing  tangents  at  the  midpoints  of  the  included 
arcs  another  circumscribed  polygon  is  formed;  the  perim- 
eter of  this  polygon  is  less  than  the  perimeter  of  the  given 
polygon  (?).  This  can  be  continued  indefinitely,  decreas- 
ing the  perimeter  of  the  polygons.  Hence,  there  can  be  no 
circumscribed  polygon  whose  perimeter  can  be  the  least  of 
all  such  polygons  ;  because,  by  increasing  the  number  of  sides, 
the  perimeter  is  lessened.  Hence,  the  circumference  must 
be  less  than  the  perimeter  of  any  circumscribed  polygon. 


BOOK  V 


225 


440.  THEOREM.  If  the  number  of  sides  of  an  inscribed  regular  poly- 
gon and  of  a  circumscribed  regular  polygon  be  indefinitely  increased, 
I.  The  perimeter  of  each  polygon  will  approach  the  circumference 
of  the  circle  as  a  limit. 

II.  The  area  of  each  polygon  will  approach  the  area  of  the  circle 
as  a  limit. 

Given:  A  circle  O,  whose  cir- 
cumference is  C  and  area  is  8 ; 
AB  and  ArBr,  sides  of  regular  cir- 
cumscribed and  inscribed  poly- 
gons, having  the  same  number  of 
sides  ;  P  and  P7,  their  perimeters ; 
K  and  E!,  their  areas. 

To  Prove :  That  if  the  number 
of  sides  be  indefinitely  increased  : 

I.  P  will  approach  C  and  P1  will  approach  C  as  limit. 
II.  K  will  approach  8  and  K1  will  approach  s  as  limit. 
Proof:  I.  The  polygons  are  similar  (?)  (425). 

.•.£.  =  .2?  (?)  (434).     Now,  if  the   number   of   sides   of 
P1      OD 

these   polygons  be  indefinitely  increased,  OD  will  approach 
OE  (?)  (437). 

/~l  77*  7-* 

Hence, — will  approach  1.    That  is,— -  will  approach  1,  or  P 
OD  P' 

and  P1  will  approach  equality ;  that  is,  they  will  approach  the 
same  constant  as  a  limit. 

But  P  >  C    and  C  >  P'  and  C  is  constant. 

Hence,  P  will  approach  C  and  P'  will  approach  C.      Q.E.D. 


III.  .fL-^L  (?)  (435).     If  the  number  of  sides  of  these 
**      OD2  _ 

polygons  be  indefinitely  increased,  OD2  will  approach  (XE2,  and 


thus 


OD 


approach  unity. 


(The  argument  continues  the  same  as  in  I.) 


226  PLANE   GEOMETRY 

441.  THEOREM.    The  circumferences  of  two  circles  are  to  each 
other  as  their  radii. 

Given  :  Two  ©  whose  radii 
are  R  and  Rr  and  circumfer- 
ences, C  and  C1  respectively.  ...••/ 

To  Prove :   C  :  Cf  =  R  :  Rf. 

Proof:  Circumscribe  regu- 
lar polygons  (having  the 

same  number  of  sides)  about  these  ©  and  let  P  and  Pf 
denote  their  perimeters.  Then,  P  :  Pf  —  R  :  R'  (?)  (434). 
Hence,  P  •  Rf  =  Pf  •  R  (?).  Now  suppose  the  number  of 
sides  of  these  polygons  to  be  indefinitely  increased, 

P   will  approach  C  (?)  (440). 
P'  will  approach  c'  (?). 
.'.  P  -  Rf  will  approach  C  •  Rf, 
and  P'  •  R  will  approach  C1  -  R. 

Hence,  C-R'  =  cr  -R  (?)  (242). 

Therefore          C  :  Cr  =  R  :  Rf  (?)  (291).  Q.E.D. 

442.  THEOREM.   The  ratio  of  any  circumference  to  its  diameter  i& 
constant  for  all  circles.     That  is,  any  circumference  divided  by  its 
diameter  is  the  same  as  any  other  circumference  divided  by  its  diam- 
eter. 

Proof:  £  =  -£.(?)  (441).   But  ^  =  |^  =  ^(?). 

0         JLV  R         Zt  R         D 

.-.£=*,  (Ax.  1). 

d     nf 

Hence,  -  =  ~  (?)  (292).  That  is,  -  =  constant.        Q.E.D. 

443.  Definition  of  TT  (pi) .      The  constant  ratio  of  a  circum- 
ference to  its  diameter  is  called  IT.      That  is, -  =  TT. 

Z) 


BOOK  V 

The  numerical  value  of  TT  =  3.141592 
(This  is  determined  in  470.) 


227 
,  approximately. 


444.  FORMULA.    Let  C  =  circumference  and  B  =  radius. 

Then,  ^-=TT  (443).      .'.  C  =  27rR   (Ax.  3). 
2* 

445.  THEOREM.   The  area  of  a  circle  is  equal  to  half  the  product  of 
its  circumference  by  its  radius. 

Given  :  O  whose  circumfer- 
ence =  (7,  area  =6',  radius  =1?. 

To  Prove  :  8  =  %  c  •  B. 

Proof  :  Circumscribe  a  reg- 
ular polygon  about  the  circle; 
denote  its  area  by  K  and  per- 
imeter by  P. 

NowJT=Jp..B(?)(436). 

Suppose  the  number  of 
sides  of  the  polygon  be  in- 
definitely increased. 

K  will  approach  S,  and  P  will  approach  C  (?). 

|-  P  •  B  will  approach  ^  C  •  B  as  a  limit  (?). 
Hence,  8  =  £  C  -  B  (?)  (242). 


.B.D. 


446.  FORMULA-    Let  8  =  area  of  0,  C  =  its  circumference,  and 
B  =  its  radius.     Then,  8  =  J  C  •  B    (445). 

Now  C=  2  TT  U  (444).     Substituting,  s  =  %  (2  TTR)  B. 

.*.   S 

Ex.  1.  Could  445  be  proved  by  inscribing  a  regular  polygon?    Why? 

Ex.  2.  The  radius  of  a  circle  is  40.    Find  the  circumference  and  area. 

Ex.  3.  The  diameter  of  a  circle  is  25.   Find  the  circumference  and  area. 

Ex.  4.  Prove  that  the  area  of  a  circle  equals  .7854  D2. 


228  PLANE   GEOMETRY 

447.   THEOREM.    The  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii,  and  as  the  squares  of  their  diameters. 

To  Prove:    8  :  81  =  R2:  Rr2  =  D2:Df2. 


And       = 

8'       (|Z/)2          l  D 

.'.  S:S'  =  R2":  It'2  =  I}2  :  D'2  (Ax.  1).  Q.E.D. 

448.  THEOREM.   The  area  of  a  sector  is  the  same  part  of  the  circle 
as  its  central  angle  is  of  360°.     (Ax.  1.) 

449.  FORMULA.   An  arc:  circum.  =  central  Z  :  360°  (244). 

.  .  arc  :  2  TTR  =  Z  :  360°. 

NOTE.   If  any  two  of  the  three  quantities,  arc,  R,  Z,  are  known,  the 
remaining  one  can  be  found  by  this  proportion. 

450.  FORMULA.    Sector  :  area  of  O  =  central  Z:  360°  (448). 

.'.  sector  :  7rRz  =  Z  :  360°. 

NOTE.    If  any  two  of  the  three  quantities,  sector,  R,  Z,  are  known,  the 
remaining  one  can  be  found  by  this  proportion. 

451.  FORMULA.    Sector  :  area  of  O  =  arc  :  circum.  (Ax.  1). 
.*.  sector  :  TT  R2  =  arc  :  2  TTR    (Ax.  6). 

.  .  sector  =  |  R  -  arc  (290). 

452.  FORMULA.   Area  of  a  segment  of  a  circle  =  area  of  the 
sector  minus*  area  of  an  isosceles  triangle. 

453.  Similar  arcs,  similar  sectors,  and 
similar  segments  are  those  which  corre- 
spond to  equal  central  angles,  in  unequal 
circles. 

Thus,  AB,  A'B',  AnBn  are  similar  arcs; 
AOB,  AfOBr,  and  A"OBrf  are  similar  sec- 
tors ;  and  the  shaded  segments  are  simi- 
lar segments. 

*  If  the  segment  is  greater  than  a  semicircle,  the  area  of  the  triangle  should 
be  added. 


BOOK  V 


229 


454.   THEOREM.   Similar  arcs  are  to  each  other  as  their  radii. 
Given :  Arcs  whose  lengths  are  a  and  a',  radii  B  and  Rf. 
To  Prove  :  a  :  a'  =  R  :  Rf. 


;  .'.a  :  a  =  B:  E. 


Q.E.D. 


455.   THEOREM.  Similar  sectors  are  to  each  other  as  the  squares  of 
their  radii. 

Given  :   Sectors  whose  areas  are  T  and  T',  radii  R  and  R'. 
To  Prove:  T  :  T'  =  R2  :  R1'2. 


...  _JL  =  JE     (?).     ...  T  :  T1  =  R2  :  R12.   (Explain.)     Q.E.D. 


456.   THEOREM.    Similar  segments  are  to  each  other  as  the  squares 
of  their  radii. 

Given:  (?).     To  Prove:  (?).  o 

Proof:  &AOB  and  A'O'B'  are 
similar  (?)  (317). 


and 


sector 


,  =  ~  (?)  (455). 
'        1'2  ^ 


sector  O'A'C'B'      R'2 

sector  PACE    _   AAOB    ,^x 
sector  O'A'C'B'     A  A'O'B' 


sector 


-  A 


n)< 


sector  O'A'C'B'  -  A  vl'o's'      A  4'o'B 

Hence,    segment  ^nc   =  ^AOB_  =  &  (Ax<  6> 

segment  A'B'C'      A  ^'o'^'      .B/2  Q.E.D. 


230  PLANE   GEOMETRY 

ORIGINAL   EXERCISES    (THEOREMS) 

1.  The  central  angle  of  a  regular  polygon  is  the  supplement  of  the 
angle  of  the  polygon. 

2.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  (if  the 
number  of  its  sides  is  odd). 

3.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 
[Draw  radii  and  apothems.] 

4.  The  sides  of  a  circumscribed  regular  polygon  are  bisected  at  the 
points  of  contact. 

5.  The  diagonals  of  a  regular  pentagon  are  equal. 

6.  The  diagonals  drawn  from  any  vertex  of  a  regular  n-gon  divide 
the  angle  at  that  vertex  into  n-2  equal  parts. 

7.  If  a  regular  polygon  be  inscribed  in  a  circle  and  another  regular 
polygon  having  the  same  number  of  sides  be  circumscribed  about  it,  the 
radius  of  the  circle  will  be  a  mean  proportional  between  the  apothem  of  the 
inner  and  the  radius  of  the  outer  polygon. 

8.  The  area  of  the  square  inscribed  in  a  sector 
whose  central  angle  is  a  right  angle  is  equal  to  half 
the   square   of  the   radius. 

[Find  z2,  the  area  of  OEDC.] 

9.  The  apothem  of  an  equilateral  triangle  is  one    O 
third  the  altitude  of  the  triangle. 

10.  The  chord  which  bisects  a  radius  of  a  circle  at  right  angles  is  the 
side  of  the  inscribed  equilateral  triangle. 

[Prove  the  central  Z.  subtended  is  120°.] 

11.  If  ABODE  is  a  regular  pentagon,  and 
diagonals  AC  and  ED  be  drawn,  meeting  at  0: 

(a)  AO  will  =  AB. 

(b)  A  O  will  be  ||  to  ED. 

(c)  ABOC  will  be  similar  to  A  BDC. 
(O  ^ACB  will  =  36°. 

(e)  A  C  will  be  divided  into  mean  and  ex- 
treme ratio  at  0. 


12.  The  altitude  of  an  equilateral  triangle  is  three  fourths  the  diameter 
of  the  circumscribed  circle. 


BOOK   V 


231 


13.  The  apothem  of  an  inscribed  regular  hexagon  equals  half  the  side 
of  an  inscribed  equilateral  triangle. 

14.  The  area  of  a  circle  is  four  times  the  area  of  another  circle 
described  upon  its  radius  as  a  diameter. 

15.  The  area  of  an  inscribed  square  is  half  the  area  of  the  circum- 
scribed square. 

16.  An  equilateral  polygon  circumscribed  about  a  circle  is  regular 
(if  the  number  of  its  sides  is  odd). 

17.  The  sum  of  the  circles  described  upon  the  legs  of  a  right  triangle 
as  diameters  is  equivalent  to  the  circle  described 

upon  the  hypotenuse  as  a  diameter. 

18.  A  circular  ring  (the  area  between  two  con- 
centric circles)  is  equivalent  to  the  circle  described 
upon  the  chord  of  the  larger  circle,  which  is  tan- 
gent to  the  less,  as  a  diameter. 

Proof  :  Draw  radii  OB,  OC.     A  OB  C  is  rt.  A  (?)  ; 
and  OC'2  -OB2  =  BC*  (?).     Etc. 

19.  If  semicircles  be  described  upon  the  three 
sides  of  a  right  triangle  (on  the  same  side  of  the 
hypotenuse),  the  sum  of  the  two  crescents  thus 
formed  will  be  equivalentto  the  areaof  the  triangle. 

f  Entire  figure  ~  \irAl?  +  crescent  B  DC  +  crescent  A  EC  (?). 


:: 


+  $7rBC2+  A  ABC   (?). 


Entire  figure-  ^ir 
Now  use  Ax.  1  ;  etc. 

20.  Show  that  the  theorem  of  No.  19  is  true  in  the  case  of  a  right  tri- 
angle whose  legs  are  18  and  24. 

21.  If  from  any  point  in  a  semicircumference  a  line  be  drawn  perpen- 
dicular to  the  diameter  and   semicircles  be  de- 

scribed on  the  two  segments  of  the  hypotenuse  as 
diameters,  the  area  of  the  surface  bounded  by 
these  three  semicircumferences  will  equal  the 
area  of  a  circle  whose  diameter  is  the  perpendicu- 
lar first  drawn. 


Proof:  Area  =  *  ,(<L±»f  -  J  .(f)*  -  i  ,(f)'= 


etc. 


22.   Show  that  the  theorem  of  No.  21  is  true  in  the  case  of  a  circle 
whose  diameter  A  B  is  25  and  AD  is  5. 


232  PLANE   GEOMETRY 

23.  If  the  sides  of  a  circumscribed  regular  polygon  are  tangent  to  the 
circle  at  the  vertices  of  an  inscribed  regular  polygon,  each  vertex  of  the 
outer  lies  on  the  prolongation  of  the  apothems  of  the  inner  polygon, 
drawn  perpendicular  to  the  several  sides. 

24.  The  sum  of  the  perpendiculars  drawn  from  any  point  within  a 
regular  n-gon  to  the  several  sides  is  constant  [=  n  •  apothem]. 

Proof :  Draw  lines  from  the  point  to  all  vertices.     Use  436  and  378. 

25.  The  area  of  a  circumscribed  equilateral  triangle  is  four  times  the 
area  of  the  inscribed  equilateral  triangle. 

26.  If  a  point  be  taken  dividing  the  diameter  of  a  circle  into  two  parts 
and  circles  be  described  upon  these  parts  as  diameters,  the  sum  of  the  cir- 
cumferences of  these  two  circles  equals  the  circumference  of  the  original 
circle. 

27.  Show  that  the  theorem  of  No.  26  is  true  in  the  case  of  a  circle 
the  segments  of  whose  diameter  are  7  and  12. 

28.  The  area  of  an  inscribed  regular  octagon  is  equal  to  the  product  of 
the  diameter  by  the  side  of  the  inscribed  square. 

29.  If  squares  be  described  on  the  six  sides  of  a  regular  hexagon 
(externally),  the  twelve  exterior  vertices  of  these  squares  will  be  the 
vertices  of  a  regular  12-gon. 

30.  If  the  alternate  vertices  of  a  regular  hexagon  be  joined  by  draw- 
ing diagonals,  another  regular  hexagon  will  be  formed.     Also  its  area 
will  be  one  third  the  original  hexagon. 

31.  Show  that  the  theorem  of  No.  18  is  true  in  the  case  of  two  con- 
centric circles  whose  radii  are  34  and  16. 

32.  In  the  same  or  equal  circles  two  sectors  are  to  each  other  as  their 
central  angles. 

33.  If  the  diameter  of  a  circle  is  10  in.  and  a  point  be  taken  dividing 
the  diameter  into  segments  whose  lengths  are  4  in.  and  6  in.,  and  on  these 
segments  as  diameters  semicircumferences  be  described  on  opposite  sides 
of  the  diameter,  these  arcs  will  form  a  curved  line  which  will  divide  the 
original  circle  into  two  parts  in  the  ratio  of  2  :  3. 

34.  If  the  diameter  of  a  circle  is  d  and  a  point  be  taken  dividing  the 
diameter  into  segments  whose  lengths  are  a  and  d  —  a,  and  on  these  seg- 
ments as  diameters  semicircumferences  be  described  on  opposite  sides  of 
the  diameter,  these  arcs  will  form  a  curved  line  which  will  divide  the 
original  circle  into  two  parts  in  the  ratio  of  a  :  d  —  a. 


BOOK  V 


288 


CONSTRUCTIONS 

457.  PROBLEM.    To  inscribe  a  square  in  a  given  circle. 

Given:  The  circle  o.  Re- 
quired  :  To  inscribe  a  square. 

Construction :  Draw  any  di- 
ameter, AB,  and  another  diame- 
ter, CD,  _L  to  AB.  Draw  AC,  BC, 
BD,  AD. 

Statement:  ACBD  is  an  in- 
scribed square.  Q.E.F. 

Proof :  A  at  O  are  =  (?). 

.-.  arcs  AC,  CB,  etc.  are  =  (?). 

.*.  ACBD  is  an  inscribed  regular  polygon  (?)  (418). 

.'.ABCD  is  a  square  (?).  Q.E.D. 

458.  PROBLEM.    To  inscribe  a  regular  hexagon  in  a  given  circle. 
Given:  (?).     Required:  (?). 

Construction:  Draw  any  ra- 
dius, AO.  At  A,  with  radius 
=  AO,  describe  arc  intersecting 
the  given  O  at  B.  Draw  AB. 

Statement :  AB  is  the  side  of 
an  inscribed  regular  hexagon. 

Proof :  Draw  BO.  A  ABO  is 
equilateral  (Const.). 

/.  A  ABO  is  equiangular  (?)  (56).     .-.  Z  AOB  =  60°(?)(115). 

.-.arc  AB —^  of  the  circumference  (J  of  360°). 

.'.  polygon   AD,  inscribed,   having   each  side  =  AB,  is  an 
inscribed  regular  hexagon  (?)  (418).  Q.E.D, 


Ex.   If  the  radius  of  a  circle  is  7  in.,  find : 
(a)  The  circumference  and  the  area. 
(6)   The  side  and  area  of  the  inscribed  square, 
(c)  The  side  and  area  of  the  inscribed  regular  hexagon. 


234 


PLANE   GEOMETRY 


459.  PROBLEM.   To  inscribe  a  regular  decagon  in  a  given  circle. 

Given:  (?).     Required:  (?). 

Construction  :  Draw  any  radius 
AO.  Divide  it  into  mean  and  ex- 
treme ratio  (by  363),  having  the 
larger  segment  next  the  center. 
Take  A  as  a  center  and  OB  as  a 
radius,  draw  an  arc  cutting  O  at 
C.  Draw  AC,  EC,  OC. 

Statement  :  AC  is  a  side  of  the 
inscribed  regular  decagon.  Q.E.F. 

Proof:   AO  :  BO  =  BO  :  AB  (Const.). 

That  is,  AO  :  AC  =  AC  :  AB  (Ax.  6).     Hence,  A  ABC  and 
AOC  have  Z  A  common  and  are  similar  (?)  (317). 

/.1st,  Z.ACB  =  ZO  (?); 
and  2d,  A  ABC  is  isosceles  (being  similar  to  A  AOC). 

Hence,  AC=BC  (?),  but  AC=BO  (?).    /.  BC=BO  (Ax.  1). 

Therefore,  Z  SCO  =  Z  O  (?)  (55). 

.-.  Z  AGO  =  2  Z  O  (Ax.  2).     Also,  Z  A  =  Z  AGO  (?)  (55). 

s.£A      =  2Zo  (Ax.  1). 
Z  O      =  1  Z  O.    Adding, 
.  2). 


Hence,  5  Z  o  =  180°  (?)  (110). 

.*.  Z  o  =  36°  (Ax.  3)  ;  that  is,  arc  AC  =  •£$  of  the  circum- 
ference (^  of  360°). 

Hence,  polygon  AE,  having  each  side  =  AC,  is  an  inscribed 
regular  decagon  (?)  (418).  Q.E.D. 


Ex.   If  the  radius  of  a  circle  is  20  in.,  find: 
(a)  The  circumference  and  area. 
(£>)   The  side  and  area  of  the  inscribed  square. 

(c)  The  side  and  area  of  the  inscribed  regular  hexagon. 

(d)  The  side  of  the  inscribed  regular  decagon.     (See  365.) 

(e)  The  area  of  sector  AOC  (fig.  459). 

(/)  The  radius  of  a  circle  containing  twice  the  area  of  this 
circle. 


BOOK  V  235 

460.  PROBLEM.   To  inscribe  a  regular  i5-gon  (pentedecagon)  in  a 
given  circle. 

Given:  (?).     Required:  (?). 

Construction :  Draw  AB,  the  side 
of  an  inscribed  hexagon,  and  AC, 
the  side  of  an  inscribed  decagon. 
Draw  BC. 

Statement :  B C  is  the  side  of  an 
inscribed  regular  15-gon.  Q.E.F. 

Proof:  Arc  JB(7=arc  AB  —  arc 
AC  =  1  —  -^  =  -±g  of  the  circum- 
ference. (Const.) 

Hence,  the  polygon  having  each  side,  a  chord,  =  BC,  is  an 
inscribed  regular  15-gon  (?)  (418).  Q.E.D. 

461.  PROBLEM.   To  inscribe  in  a  given  circle: 

I.  A  regular  8-gon,  a  regular  i6-gon,  a  regular  32-gon,  etc. 
II.    A  regular  i2-gon,  24-gon,  etc. 

III.    A  regular  so-gon,  6o-gon,  etc. 

Construction:  I.   Inscribe  a  square ;  bisect  the  arcs ;  draw 
chords.     Statement:   (?).     Proof:   (?).     (See  422.)    Etc. 

II.  Inscribe  a  regular  hexagon ;  bisect  the  arcs.     Etc. 
III.  Inscribe  a  regular  15-gon,  etc. 

462.  PROBLEM.   To  inscribe  an  equilateral  triangle  in  a  circle. 

Construction:  Join  the  alternate  vertices  of  an   inscribed 
regular  hexagon.     Proof:   (?)     (See  421.) 

463.  PROBLEM.    To  inscribe  a  regular  pentagon  in  a  given  circle. 

464.  PROBLEM.   To  circumscribe  a  regular  polygon  about  a  circle. 

Construction :  Inscribe  a  polygon  having  the  same  num- 
ber of  sides.     At  the  several  vertices  draw  tangents. 

Statement:  (?).     Proof:  (?).     (See  419.) 


236 


PLANE   GEOMETRY 


FORMULAS 
465.    Sides  of  inscribed  polygons. 

1.  Side  of  inscribed  equilat- 
eral triangle  =  R  V3. 

Proof :  Z  ACS  is  a  rt.  Z  (?). 
AB  =  2  R  and  CB  =  R  (?)• 


2.  Side   of  inscribed  square 
=  tfV2. 

Proof:  Use  fig.  of  457. 

3.  Side  of  inscribed  regular  hexagon  =  R  (?). 

4.  Side  of  inscribed  regular  decagon  =  %R  (V5  - 1).  (365.) 

466.  Sides  of  circumscribed  polygons. 

1.  Side   of   circumscribed 

rv  B  e- 

equilateral  A  =  2  R  V3. 

Proof  :    Z.  DAB  =  Z.  DBA 

=  ZD  =  60°(?). 
/.A  ABD  is  equilateral. 
AD  =  AB  =  R_V3  (?). 
.'.DF=2R  V3. 

2.  Side   of   circumscribed 
square=  2  B  (?). 

3.  Side  of    circumscribed 
regular  hexagon  =  |  R  V3. 
(Explain.) 

467.  In  equilateral  triangle,  apothem  =  |  P. 
Proof:   Bisect  arc  AC  at  H.     Draw  OA,  OC, 
Figure  AOCH  is  a  rhombus.     (Explain.) 
0^=i  OH  =  IE  (?)  (141). 


BOOK  V 


237 


468.  PROBLEM.  In  a  circle  whose  radius  is  R  is  inscribed  a  regu- 
lar polygon  whose  side  is  s ;  to  find  the  formula  for  the  side  of  an 
inscribed  regular  polygon  having  double  the  number  of  sides. 

Given :  AB  =  s,  a  side  of  an  in- 
scribed regular  polygon  in  O 
whose  radius  is  R;  C,  the  mid- 
point of  arc  AB]  chord  AC. 

Required :  To  find  the  value  of 
AC,  the  side  of  a  regular  polygon 
having  double  the  number  of 
sides  and  inscribed  in  the  same 
circle. 

Construction:  Draw  radii  OA 
and  OC. 

Computation :  OC  bisects  AB  at  right  A  (?)  (70). 

In  rt.*  A  AON,  0  is  an  acute  Z.     Hence  in  A  AOC, 
AC*  =  ol2  +  oc2  -  2  .  oc  •  ON  (?)  (346). 

[0  =  R,  00=  R,   ON  = 

.'.  AC2  =  2  #2  -  2  R  •  1  V4  R2  -  s2  (Ax.  6)  ; 

AC  = 


or 


Q.E.F. 


469.  FORMULA.    If  R  =  1,  and  given  side  =  s,   the  side    of   a 
regular  polygon  having  twice  as  many  sides  =  \2-V4  —  s2. 


Ex.  1.   If  the  radius  of  a  circle  is  4,  find : 

(a)  The  side  of  the  inscribed  equilateral  triangle. 

(b)  The  side  of  the  circumscribed  equilateral  triangle. 

(c)  The  side  of  the  inscribed  square. 

(d)  The  side  of  the  circumscribed  square. 

(e)  The  side  of  the  inscribed  regular  hexagon. 

(y)  The  side  of  the  circumscribed  regular  hexagon. 
(g)  The  apothem  of  the  inscribed  equilateral  triangle. 
(Ji)  The  apothem  of  the  inscribed  regular  hexagon. 
(i)  The  side  of  the  inscribed  regular  dodecagon  (468). 
(/)  The  side  of  the  inscribed  regular  octagon. 


238 


PLANE   GEOMETRY 


470.   PROBLEM.   To  find  the  approximate  numerical  value  of  ir. 
Given :  A  circle  whose  diameter  =  D  and  circumference  =  c. 
Required :  The  value  of  TT,  that  is,  the  value  of  C  -j-  D. 
Method:  1.  We  may  select  a  O  of  any  diameter  (442). 
Hence,  for  simplicity,  we  take  the  O  in  which  D  =  2  ;  /.  R  =  1. 

2.  We  compute  the  perimeter  of  some  inscribed  regular 
polygon  (by  465). 

3.  We  compute  the  length   of   a  side   of   the   inscribed 
regular  polygon  having  double  the  number  of  sides,  by  the 

formula  «  =  \2  —  V4  —  s2  (469).      From  this  we   can   find 
the  perimeter  of  this  polygon. 

4.  Using  the  side  of  this  polygon  as  known,  we  compute, 
by  the  same  formula,  a  side  of  the  inscribed  regular  polygon 
having  still  double  the  number  of  sides.      Hence  its  perim- 
eter can  be  found. 

5.  By  continuing  this  process  we  may  approximate  the 
value  of  the  circumference  (440,  I). 

6.  Thus  we  can  find  the  value  of  C  -J-  D,  or  TT. 
Computation :  1.  Assume  E  =  1. 

2.  Consider  the  regular  hexagon  and  let  s6  represent  its 
side  and  P6  its  perimeter.     Then  s6  =  1  and  P6  =  6  (?) . 

3.  Then,    a    side    of    the    inscribed   regular    12-gon     is 

/2-V4^1=  0.5176381,  and  P12  =  6.2116572. 


'12 

4.  Thus  we  may  find  s24=  0.2610524  and  P24=  6.2652576. 

5.  By  continuing,  s3072  =  0.002045,  and  P3072  =  6.283184. 

6.  /.  it  is  evident  that  C=  6.283184,  approximately. 


But,  ir=-  (?). 


7T   = 


6.283184 


=  3.141592 +.    Q.E.F. 


This  calculation  is  tabulated  for  reference. 


*6=1,  .-./>„=  8. 

s12  =  0.517638,  .-.  P12  =  6.211657. 

s24  =  0.261052,  .-.  P24  =  6.265257. 

$48  =  0.130806,  .-.  P48  =  6.278700. 

fe  =  0.065438,  .-.  P96  =  6.282063. 


sm  =  0.032723,  .-.  P 

s384  =  0.016362,  .-.  P 

sm  =  0.008181,  .-.  P 

sls36=  0.004091,  .-.  P1 

$3072=  0.002045,  .-.  P3 


=  6.282904. 
=  6.283115. 
=  6.283169. 
=  6.283180. 
=  6.283184. 


BOOK  V  239 

ORIGINAL   EXERCISES   (NUMERICAL) 

MENSURATION  OF  REGULAR  POLYGONS  AND  THE  CIRCLE 

1.  Find  the  angle  and  the  central  angle  of: 

(i)  a  regular  pentagon ;  (ii)  a  regular  octagon ;    (iii)   a  regular  do- 
decagon; (iv)  a  regular  20-gon. 

2.  Find  the  area  of  a  regular  hexagon  whose  side  is  8. 

3.  Find  the  area  of  a  regular  hexagon  whose  apothem  is  4. 

4.  In  a  circle  whose  radius  is  10  are  inscribed  an  equilateral  tri- 
angle, a  square,  and  a  regular  hexagon.     Find  the  perimeter,  apothem, 
and  area  of  each. 

5.  About  a  circle  whose  radius  is  10  are  circumscribed  an  equilat- 
eral triangle,  a  square,  and  a  regular  hexagon.     Find  the  perimeter  and 
area  of  each. 

6.  Find  the  circumference  and  area  of  a  circle  whose  radius  is  5 
inches.     [Use  TT  =  3^.] 

7.  Find  the  circumference  and  area  of  a  circle  whose  diameter  is 
42  centimeters. 

8.  The  radius  of  a  certain  circle  is  9  meters.    What  is  the  radius  of  a 
second  circle  whose  circumference  is  twice   as  long  as  the  first?   Of  a 
third  circle  whose  area  is  twice  as  great  as  the  first  ? 

9.  If  the  circumference  of  a  circle  is  55  yards,  what  is  its  diameter  ? 

10.  If  the  area  of  a  circle  is  113£  square  meters,  what  is  its  radius? 

11.  In  a  circle  whose  radius  is  35  there  is  a  sector  whose  angle  is  40°. 
Find  the  length  of  the  arc  and  the  area  of  the  sector. 

12.  The  area  of  a  circle  is  6^  times  the  area  of  another.     If  the 
radius  of  the  smaller  circle  is  12,  what  is  the  radius  of  the  larger  circle  ? 

13.  If  the  angle  of  a  sector  is  72°  and  its  arc  is  44  inches,  what  is 
the  radius  of  the  circle  ?  What  is  the  area  of  the  sector  ? 

14.  In  a  circle  whose  radius  is  7  find  the  area  of  the  segment  whose 
central  angle  is  120°.     [See  451.] 

15.  If  the  radius  of  a  circle  is  4  feet,  what  is  the  area  of  a  segment 
whose  arc  is  60°  ?  of  a  segment  whose  arc  is  a  quadrant? 

16.  Find  the  area  of  the  circle  inscribed  in  a  square  whose  area  is  75. 


240  PLANE   GEOMETRY 

17.  Find  the  area  of  an  equilateral  triangle  inscribed  in   a  circle 
whose  area  is  441  TT  square  meters. 

18.  If  the  length  of  a  quadrant  is  8  inches,  what  is  the  radius  ? 

19.  Find  the  length  of  an  arc  subtended  by  the  side  of  an  inscribed 
regular  15-gon  if  the  radius  is  4f  inches. 

20.  The  side  of  an  equilateral  triangle  is  10.    Find  the  areas  of  its  in- 
scribed and  circumscribed  circles. 

21.  Find  the  perimeter  and  area  of  a  segment  whose^  chord  is  the 
side  of  an  inscribed  regular  hexagon,  if  the  radius  of  a  circle  is  5£. 

22.  A  circular  lake  9  rods  in  diameter  is  surrounded  by  a  walk  \  rod 
wide.     What  is  the  area  of  the  walk  ? 

23.  A  locomotive  driving  wheel   is  7  feet  in  diameter.     How  many 
revolutions  will  it  make  in  running  a  mile? 

24.  What  is  the  number  of  degrees  in  the  central  angle  whose  arc  is 
as  long  as  the  radius  ? 

25.  Find  the  side  of  the  square  equivalent  to  a  circle  whose  diameter 
is  4.2  meters. 

26.  Find  the  radius  of  that  circle  equivalent  to  a  square  whose  side 
is  5.5  inches. 

27.  Find  the  radius  of  the  circumference  which  divides  a  given  circle 
whose  radius  is  10^  into  two  equal  parts. 

28.  Three  equal  circles  are  each  tangent  to  the  other  two  and  the 
diameter  of  each  is  40  feet.     Find  the  area  between  these  circles. 

[Required  area  =  area  of  an  eq.  A  minus  area  of  three  sectors.] 

29.  Find  the  area  of  the  three  segments  of  a  circle  whose  radius  is 
5\/3,  formed  by  the  sides  of  the  inscribed  equilateral  triangle. 

30.  If  a  cistern  can  be  emptied  in  5  hours  by  a  2-inch  pipe,  how  long 
will  be  required  to  empty  it  by  a  1-inch  pipe  ? 

31.  Find  the  side,  apothem,  and  area  of  a  regular  decagon  inscribed 
in  a  circle  whose  radius  is  6  feet. 

32.  What  is  the  area  of  the  circle  circumscribed  about  an  equilateral 
triangle  whose  area  is  48  V3? 

33.  The  circumferences  of  two  concentric  circles  are  40  inches  and  50 
inches.     Find  the  area  of  the  circular  ring  between  them. 

34.  A  circle  has  an  area  of  80  square  feet.     Find  the  length  of  an 
arc  of  80°. 


BOOK  V 


241 


35.  Find  the  angle  of  a  sector  whose  perimeter  equals  the  circum- 
ference. 

36.  Find  the  angle  of  a  sector  whose  area  is  equal  to  the  square 
of  the  radius. 

37.  Find  the  area  of  a  regular  octagon  inscribed  in  a  circle  whose 
radius  is  20. 

[Inscribe  square,  then  octagon.     Draw  radii  of  octagon.     Find  area  of 
one  isosceles  A  formed,  whose  altitude  is  half  the  side  of  the  square.] 

38.  A  rectangle  whose  length  is  double  its  width,  a  square,  an  equi- 
lateral triangle,  and  a  circle  all  have  the  same  perimeter,  namely  132 
meters.     Which  has  the  greatest  area?  the  least? 

39.  Through  a  point  without  a  circle  whose  radius  is  35  inches  two 
tangents  are  drawn,  forming  an  angle  of  60°.     Find  the  perimeter  and 
area  of  the  figure  bounded  by  the  tangents  and  their  smaller  intercepted  arc. 

40.  In  a  circle  whose  radius  is  12  are  two  parallel  chords  which  sub- 
tend arcs  of  60°  and  90°  respectively.     Find  the  perimeter  and  area  of 
the  figure  bounded  by  these  chords  and  their  intercepted  arcs. 

41.  A  quarter  mile  race  track  is  to  be  laid  out,  having  parallel  sides 
but  semicircular  ends  whose  radius  is  105  feet.     Find  the  length  of  the 
parallel  sides. 

42.  If  the  diameter  of  the  earth  is  7920  miles,  how  far  at  sea  can  the 
light  from  a  lighthouse  150  feet  high  be  seen  ? 

43.  The  diameter  of  a  circle  is  18  inches.     Find  the  area  of  the  figure 
between  this  circle  and  the  circumscribed  equilateral  triangle. 

44.  How  far  does  the  end   of    the   minute 
hand  of  a  clock  move  in  20  minutes,  if  the  hand 
is  3 1  inches  long  ? 

45.  The  diameter  of  a  circle   is   16   inches. 
What  is  the  area  of  that  portion  of  the  circle 
outside  the  inscribed  regular  hexagon  ? 

46.  Using  the  vertices  of  a  square  whose  side 
is  12,  as  centers,  and  radii  equal  to  4,  four  quad- 
rants are  described  within  the  square.     Find  the 
perimeter  and  area  of  the  figure  thus  formed. 

47.  Using  the  four  vertices  of  a  square,  whose 
side  is  12,  as   centers  and  radii  equal  to  6,  four 
arcs  are  described  without  the  square  (see  figure). 
Find  the  perimeter  and  area  of  the  figure  bounded 
by  these  four 


242 


PLANE   GEOMETRY 


48.  Using  the  vertices  of  an  equilateral  triangle,  whose  side  is  16, 
as  centers  and  radii  equal  to  8,  three  arcs  are  described  within  the  tri- 
angle.    Find  the  perimeter  and  area  of  the  figure  bounded  by  these  arcs. 
Do  the  same  if  the  three  arcs  are  described  without  the  triangle  (terminat- 
ing in  the  sides,  in  each  case). 

49.  Using  the  vertices  of  a  regular  hexagon,  whose  side  is  20,  as 
centers  and  radii  equal  to  10,  six  arcs  are  described  within  the  hexagon. 
Find  the  perimeter  and  area  of  the  figure  bounded  by  these  arcs.     Do 
the  same  if  the  six  arcs  are  described  without  the  hexagon  (terminating 
in  the  sides,  in  each  case). 

50.  If  semicircumferences  be  described  within 
a  square,  whose  side  is  8  inches,  upon  the  four 
sides  as  diameters,  find  the  areas  of  the  four 
lobes  bounded  by  the  eight  quadrants.      Find 
the  area  of  any  one. 

In  the  following  exercises  let  n  =  the  number 
of  sides  of  the  regular  polygon  ;  s  =  the  length 
of  its  side  ;  r  =  its  apothem  ;  R  =  its  radius  ; 
K  =  its  area. 


51.    If  n  =  3,  show  that  s  =  tf  V3  ;  r=*R;  K  = 


52.    If  n  =  4,  show  that  s  =  R  V2  =  2r;  K  =  2 

2r  V3 


53.    If  n  =  6,  show  that  s  =  R  = 


3 


54.  If  n  =  -8,  show  that  s  =  R^2  -  \/2  =  2r  (V2- 1) ;  r =^ J2  +  V2  ; 

R  =  r^4:-2  V2;  ^=  2  .R2  \/2  ^  8r2  (V2- 1). 

55.  If  n  =  10,  show  that  s  =  —  (\/5-l);    r=-JlO  +  2V5. 

56.  If  n  =    5,  show  that  s  =  — 


57.   If  n  =  12  show  that  s  =  R*\J2-V3  =  2  r(2-V3);  R  =  2  r 


BOOK  V  243 


58.  The  apothem  of  a  regular  hexagon  is  18  V3  inches.     Find  its 
side  and  area.    .Find  the  area  of  the  circle  circumscribed  about  it. 

59.  What  is  the  radius  of  a  circle  whose  area  is  doubled  by  increas- 
ing the  radius  10  feet  V 

60.  If  an  8-inch  pipe  will  fill  a  cistern  in  3  hours  20  minutes,  how 
long  will  it  require  a  2-inch  pipe  to  fill  it  ? 

61.  The  radius  of  a  circle  is  12  meters.     Find  : 
(a)    The  area  of  the  inscribed  square. 

(i)  The  area  of  the  inscribed  equilateral  triangle. 

(c)  The  area  of  the  inscribed  regular  hexagon. 

(rf)  The  area  of  the  inscribed  regular  dodecagon. 

(e)  The  area  of  the  circumscribed  square. 

(/)  The  area  of  the  circumscribed  equilateral  triangle. 

(<7)  The  area  of  the  circumscribed  regular  hexagon. 

Qi)  The  area  of  the  circumscribed  regular  dodecagon. 

62.  The  radius  of  a  circle  is  18.     Find  : 

(a)  The  side  and  apothem  of  the  inscribed  square. 

(b)  The  side  and  apothem  of  the  inscribed  equilateral  triangle. 

(c)  The  side  and  apothem  of  the  inscribed  regular  hexagon, 
(e?)  The  area  of  the  inscribed  square. 

(e)    The  area  of  the  inscribed  equilateral  triangle. 
(/)   The  area  of  the  inscribed  regular  hexagon. 
(g)    The  area  of  the  inscribed  regular  octagon. 
(h)    The  area  of  the  circumscribed  regular  hexagon. 

63.  Prove  that  the  area  of  an  inscribed  regular  hexagon  is  a  mean  pro- 
portional between   the   areas   of  the  inscribed  and  the   circumscribed 
equilateral  triangles.     [Find  the  three  areas  in  terms  of  jR.] 

64.  A  B  is  one  side  of  an  inscribed  equilateral  triangle,  and  C  is  the 
midpoint  of  AB.     If  AB  be  prolonged  to  0  making  BO  equal  to  BC, 
and  OT\)&  drawn  tangent  to  the  circle  at  T7,  OT  will  be  f  the  radius. 

65.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle 
all  have  the  same  area,  namely  5544  sq.  ft.     Which  figure  has  the  least 
perimeter  ?  the  greatest  ? 

66.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle  all 
have  the  same  perimeter,  namely  396  in.     Find  their  areas  and  compare. 

67.  The  circumferences  of  two  concentric  circles  are  330  and  440  in. 
respectively.     Find  the  radius  of  another  circle  equivalent  to  the  ring 
between  these  two  circumferences. 


244 


PLANE   GEOMETRY 


ORIGINAL   CONSTRUCTIONS 

1.  To  circumscribe  a  regular  hexagon  about  a  given  circle. 

2.  To  circumscribe  an  equilateral  triangle  about  a  given   circle. 

3.  To  circumscribe   a   regular  decagon    about   a    given    circle;    a 
regular  16-gon  ;  a  regular  24-gon  ;  a  square. 

4.  To   construct  an  angle  of  36°;  of  18°;  of  72°;  of  24°;  of  6°;  of 
48°;  of  96°. 

5.  To  construct  a  regular  hexagon  upon  a  given  line  as  a  side. 

6.  To  construct  a  regular  decagon  upon  a  given  line  as  a  side. 

7.  To  construct  a  regular  octagon  upon  a  given  line  as  a  side. 

8.  To  construct  a  regular  pentagon  upon  a  given  line  as  a  side. 

9.  To  construct  a  square  which  shall  have  double  the  area  of  a 
given  square. 

10.  To  inscribe  in  a  given 
circle  a  regular  polygon  simi- 
lar to  a  given  regular  polygon. 

Construction :  From  the 
center  of  the  polygon  draw 
radii.  At  the  center  of  the 
circle  construct  A-  =  these  cen- 
tral A  of  the  polygon.  Draw  chords.  Etc. 

11.  To  construct  a  regular  pentagon  which  shall  have  double  the  area 
of  a  given  regular  pentagon. 

12.  To  construct  a  circumference  equal  to  the  sum  of  two  given  cir- 
cumferences. 

13.  To  construct  a  circumference  which  shall  be  three  times  a  given 
circumference. 

14.  To  construct  a  circumference  equal  to  the  difference  of  two  given 
circumferences. 

15.  To  construct  a  circle  whose  area  shall  be  five  times  a  given  circle. 

16.  To  construct  a  circle  equivalent  to  the  sum  of  two  given  circles ; 
another,  equivalent  to  their  difference. 

17.  To  construct  a  circle  whose  area  shall  be  half  a  given  circle. 

18.  To  bisect  the  area  of  a  given  circle  by  a  concentric  circumference. 

19.  To  divide  a  given  circumference  into  two  parts  which  shall  be  in 
the  ratio  of  3:7;  into  two  other  parts  which  shall  be  in  the  ratio  of 
5:7;  into  still  two  other  parts,  in  the  ratio  of  8  : 7. 


BOOK  V  245 


MAXIMA   AND  MINIMA 

471.  Of   geometrical   magnitudes  which   satisfy   a   given 
condition   (or  given  conditions)    the  greatest  is  maximum, 
and  the  least  is  minimum. 

Thus,  of  all  chords  that  can  be  drawn  through  a  given  point  within 
a  circle,  the  diameter  is  the  maximum,  and  the  chord  perpendicular  to 
the  diameter  at  the  point  is  the  minimum. 

Isoperimetric  figures  are  figures  having  equal  perimeters. 

472.  THEOREM.   Of  all  triangles  having  two  given  sides,  that  in 
which  these  sides  form  a  right  angle  is  the  maximum. 

Given:    A  ABC  and  A  ABD 
having     AB     common,     and  D 
AC  =  AD,  but  Z  CAB  a  rt.  Z 
and  Z  DAB  not  a  right  Z. 
To  Prove :  A  ABC  >  A  ABD. 
Proof  :    Draw  altitude  DE. 
Now  AD  >  DE  (?).      .'.  AC  >  DE  (Ax.    6). 
Multiply  each  member  by  ^  AB. 
Then  J  AB  •  AC  >  J  AB  •  DE  (?). 
Now  £  AB  -  AC  =  area  A  ABC  (?), 
and        l  AB  •  DE  ==  area  A  ABD  (?). 

Therefore,  A  ABC  >  A  ABD  (Ax.  6).        Q.E.D. 

This  theorem  may  be  stated  thus  :  Of  all  triangles  having 
two  given  sides,  that  triangle  whose  third  side  is  the  diameter 
of  the  circle  which  circumscribes  it  is  the  maximum. 

Therefore,  Of  all  n-gons  having  n  —  ~L  sides  given,  that  poly- 
gon whose  nth  side  is  the  diameter  of  a  circle  which  circum- 
scribes the  polygon  is  the  maximum. 

Ex.  1.  Of  all  parallelograms  having  two  adjacent  sides  given,  the 
rectangle  is  the  maximum. 

Ex.  2.  Of  all  lines  that  can  be  drawn  from  an  external  point  to  a  cir- 
cumference, which  is  the  maximum?  the  minimum  ? 


246 


PLANE   GEOMETRY 


473.  THEOREM.   Of  all  isoperimetric  triangles  having  the  same 
base  the  isosceles  triangle  is  the  maximum. 

Given  :  A  ABC  and  ABD 
isoperimetric,  having  the 
same  base,  AB,  and  A  ABG 
isosceles. 

To  Prove  : 
A  ABC  >  A  ABD. 
Proof  :    Prolong  AC  to  E9 
making  CE  =  AC,  and  draw 
BE.      Using  D  as  a   center 
and  BD  as  a  radius,  describe 
an  arc  cutting  EB  prolonged, 
at  F.   Draw  CG  and  DH  \\  to 
AB,  meeting  EF  at  G  and  H  respectively.     Draw  AF. 

Now,  using  C  as  a  center  and  AC  or  BC  or  EC  as  a  radius, 
the  circle  described  will  pass  through  A,  B,  and  E  (Hyp. 
and  Const.).  /.  Z  ABE  =  rt.  Z  (?). 

That  is,  AB  is  _L  to  EF.     Hence,  CG  and  DH  are  J_  to  EF(?~). 
AC  +  CE  =  AC  +  CB  =  AD  +  DB  =  AD  +  DF  (Hyp.   and 
Const.). 

That  is,  AE  =  AD  +  DF  (Ax.    1). 
But  AD  +  DF>AF  (?).      .'.  AE  >  AF  (Ax.  6). 
.-.  BE  >  BF  (?)   (90),  and  ±  BE  >  ±  BF  (?). 
Now,  BG  =  ±  BE  and  BH  =  ±  BF  (?)  (73,  Cor.). 
/.  BG  >  £#  (Ax.  6). 
Multiply  each  member  by  \  AB. 
Then,  ±  AB  •  BG>±  AB  •  BH(?). 
But,    %  AB  •  BG  =  area  A  ABC  (?), 
and          J  ^#  -  BH  =  area  A  ^4£D  (?). 
.'.  A  ABC  >  A  ABD   (Ax.  6). 

474.  THEOREM. 
is  the  maximum. 

[Any  side  may  be  considered  the  base.] 


Q.E.D. 
Of  isoperimetric  triangles  the  equilateral  triangle 


BOOK  V 


247 


475.  THEOREM.   Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides  the  maximum  is  equilateral. 

Given  :  Polygon  AD,  the  max- 
imum of  all  polygons  having 
the  same  perimeter  and  the  same 
number  of  sides. 

To  Prove  :    AB  =  EC  =  CD  =  F< 
DE  =  etc. 

Proof  :  Draw  AC  and  suppose 
AB  not  =  BC. 

On    AC    as     base,     construct 
A  ACM  isoperimetric  with  A  ABC  and  isosceles;  that  is,  make 
AM  =  CM.      Then  A  ACM  >  A  ABC    (?)    (473). 
1  Add  to  each  member,  the  polygon  ACDEF. 

.'.  polygon  AMCDEF  >  polygon  AD  (?). 

But  the  polygon  AD  is  maximum  (Hyp.). 

/.  AB  cannot  be  unequal  to  BC  as  we  supposed  (because 
that  results  in  an  impossible  conclusion). 

Hence,  AB  =  BC.    Likewise  it  is  proved  that  BC  =  CD  —  etc. 

Q.B.D. 

476.  THEOREM.    Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides  the  equilateral  polygon  is  maximum. 

Proof  :  Only  one  such  polygon  is  maximum,  and  the  maxi- 
mum is  equilateral  (475). 

Only  one  such  polygon  is  equilateral,  hence  the  equilateral 
polygon  and  the  maximum  polygon  are  the  same.  Q.E.D. 


Ex.  1.   Of  isoperimetric  triangles,  the  maximum  is  equilateral. 

Ex.  2.  Of  all  right  triangles  that  can  be  constructed  upon  a  given 
hypotenuse,  which  is  maximum  ?  Why  ? 

Ex.  3.  Of  all  triangles  having  a  given  base  and  a  given  vertex-angle, 
the  isosceles  is  the  maximum. 

Ex.  4.  Of  all  mutually  equilateral  polygons,  that  which  can  be  in- 
scribed in  a  circle  is  the  maximum. 


248 


PLANE   GEOMETRY 


477.  THEOREM.  Of  isoperimetric  regular  polygons,  the  polygon 
having  the  greatest  number  of  sides  is  maximum. 

Given :  Equilateral  A  ABC 
and  square  /S,  having  the  same 
perimeter. 

To  Prove :  Square  S>AABC. 

Proof :  Take  D,  any  point  in 
BC,  and  draw  AD.  On  AD  as 
base,  construct  isosceles  A  ADE,  isoperimetric  with  A  ABD. 

Now  A  AED  >  A  ABD  (?)  (473). 

Adding  A  ADC  to  each  member,  AEDC  >  A  ABC  (?). 

AEDC  is  isoperimetric  with  A  J.BCand  s  (Hyp.  and  Const.). 

Hence,  S  >  AEDC  (?)  (476). 

Therefore  8  >  A  ABC  (?)  (Ax.  11). 

Similarly  we  may  prove  that  an  isoperimetric  regular  pen- 
tagon is  greater  than  8 ;  and  an  isoperimetric  regular  hexa- 
gon is  greater  than  this  pentagon,  etc. 

Therefore,  the  regular  polygon  having  the  greatest  num- 
ber of  sides  is  maximum.  Q.E.D. 


478.  THEOREM.   Of  all  isoperimetric  plane  figures  the  circle  is  the 
maximum, 

479.  THEOREM.    Of  equivalent  regular  polygons  the  perimeter  of 
the  polygon  having  the  greatest  number  of  sides  is  the  minimum. 


Given :   Any  two  equivalent  regular  polygons,  A  and  5,  A 
having  the  greater  number  of  sides. 


BOOK  V  249 

To  Prove  ;  The  perimeter  of  A  <  the  perimeter  of  B. 
Proof:  Construct    regular    polygon    s,   similar  to  B  and 
Ifeoperirnetric  with  A. 

Then  A  >  8  (477),  but  A  =c=  B  (?).    .-.  B  >  s  (?)  (Ax.  6). 

Hence,  the  perimeter  of  B  >  perimeter  of  S  (390). 

But,  the  perimeter  of  /S  =  perimeter  of  A  (?). 

.'.  perimeter  of  B  >  perimeter  of  A  (Ax.  6). 

That  is,  the  perimeter  of  A<  the  perimeter  of  B.       Q.E.D. 

480.  THEOREM.   Of  all  equivalent  plane  figures  the  circle  has  the 
minimum  perimeter. 

ORIGINAL   EXERCISES 

1.  Of  all  equivalent  parallelograms  having  equal  bases  the  rectangle 
has  the  minimum  perimeter. 

2.  Of  all  lines  drawn  between  two  given  parallels  (terminating  both 
ways  in  the  parallels),  which  is  the  minimum  ?    Prove. 

3.  Of  all  straight  lines  that  can  be  drawn  on  the  ceiling  of  a  room 
12  feet  long  and  9  feet  wide,  what  is  the  length  of  the  maximum? 

4.  Find  the  areas  of  an  equilateral  triangle,  a  square,  a  regular  hex- 
agon, and  a  circle,  the  perimeter  of  each  being  264  inches.     Which  is 
maximum  ?    What  theorem  does  this  exercise  illustrate  ? 

5.  Find  the  perimeters  of  an  equilateral  triangle,  a  square,  a  regular 
hexagon,  and  a  circle,  if  the  area  of  each  is  1386  square  feet.     Which 
perimeter  is  the  minimum  ?    What  theorem  does  this  exercise  illustrate  ? 

6.  Of  isoperimetric  rectangles  which  is  maximum? 

7.  To  divide  a  given  line  into  two  parts  such  D 
that  their  product  (rectangle)  is  maximum. 

8.  Of  all    equivalent  triangles   having   the 
same  base  the  isosceles  triangle  has  the  minimum 
perimeter. 

To  Prove:    The   perimeter   of  A  ABC  <  the 
perimeter  of  A  A  B'  C.  4 

Proof:  AD<AB'  +  £'C;  etc.  A 


9.   Of  all  rectangles  inscribed  in  a  circle  which  is  maximum  ?  Prove. 

10.  Of  all  rectangles  inscribed  in  a  semicircle  which  is  maximum? 

Prove. 

» 

11.  Of  all  equivalent  rectangles,  the  square  has  the  minimum  per- 
imeter. 


250  PLANE    GEOMETRY 

12.  Of  all  triangles  having  a  given  base  and  a  given  vertex-angle,  the 
isosceles  triangle  has  the  maximum  area. 

13.  Of  all  triangles  having  a  given  altitude  and  a  given  vertex-angle, 
the  isosceles  triangle  is  the  minimum. 

14.  Of  all  triangles  that  can  be  inscribed  in  a  given  circle  the  equi- 
lateral triangle  has  the  maximum  area. 

15.  The  cross  section  of  a  bee's  cell  is  a  regular  hexagon.    Would  this 
be  the  most  economical  for  the  bee,  if  one  cell  in  a  hive  were  all  he  were 
to  fill  (that  is,  would  he  use  the  least  wax)  ?    Considering  also  the  adjoin- 
ing cells,  does  the  form  of  the  regular  hexagon  require  the  least  wax? 
Explain.     Does  it  also  permit  the  storing  of  the  most  honey  ?    Why? 

16.  Prove  that  a  regular  hexagon  is  greater  than  an  isoperirnetric 
square,  by  the  method  employed  in  477. 

17.  Answer  the  questions  of  exercise  65  on  page  243,  without  any 
computation.     Give  reasons. 

18.  Compare  the  areas  of  the  figures  mentioned  in  exercise  66,  page 
243,  without  performing  any  computation. 


0 

SOLID   GEOMETRY 


BOOK  VI 

LINES,  PLANES,  AND  ANGLES  IN  SPACE 

481.  A  solid  is  any  limited  portion  of  space.     The  boun- 
daries of  a  solid  are  surfaces. 

A  plane  is  a  surface  in  which,  if  any  two  points  are  taken, 
the  straight  line  connecting  them  lies  wholly  in  that  surface. 

Solid  Geometry  is  a  science  that  treats  of  magnitudes,  all  of 
which  are  not  in  the  same  plane. 

482.  The 'intersection  of  two  surfaces  is  the  line,  or  the 
lines,  all  of  whose  points  lie  in  both  surfaces.     The  inter- 
section of  a  line  and  a  surface  is  the  point,  or  points,  com- 
mon to  both  the  line  and  the  surface.     The  foot  of  a  line 
intersecting  a  plane  is  their  point  of  intersection. 

483.  A  straight  line  is  perpendicular  to  a  plane  if  the  line 
is  perpendicular  to  every  straight  line  in  the  plane  drawn 
through  its  foot. 

A  normal  is  a  straight  line  perpendicular  to  a  plane. 

484.  A  straight  line  is  parallel  to  a  plane  if  the  line  and 
the    plane    never    meet,    when    indefinitely   extended.     A 
straight  line  is  oblique  to  a  plane  if  it  is  neither  perpendicular 
nor  parallel  to  the  plane.     Two  planes  are  parallel  if  they 
never  meet  when  indefinitely  extended. 

261 


252  SOLID  GEOMETRY 

485.  The  projection  of  a  point  upon  ^a  plane  is  the  foot  of 
the  perpendicular  from  the  point  to  the  plane. 

The  projection  of  a  line  upon  a  plane  is  the  line  formed  by 
the  projections  of  all  the  points  of  the  given  line. 

486.  A  plane  is  determined  if  its  position  is  fixed  and  if 
that  position  can  be  occupied  bj  only  one  plane. 


PRELIMINARY   THEOREMS 

487.  THEOREM.     If  two  points  of  a  straight  line  are  in  a  plane, 
the  whole  line  is  in  the  plane.     [Def.  481.] 

488.  THEOREM.     A  straight  line  can  intersect  a  plane  in  only  one 
point.     [See  487.] 

489.  THEOREM.     If  a  line  is  perpendicular  to  a  plane,  it  is  perpen- 
dicular to  every  line  in  the  plane  drawn  through  its  foot.     [See  483.] 

490.  THEOREM.     Through  one  straight  line  any  number  of  planes 
may  be  passed. 

Because,  if  we  consider  a  plane  con- 
taining a  line  AB  to  revolve  about  AB, 
it  may  occupy  an  indefinitely  great  num- 
ber  of  positions.  Each  of  these  will  be 
a  different  plane  containing  AB. 

491.  THEOREM.     Through  a  fixed  straight  line  and  an  external 
point  a  plane  can  be  passed. 

Because,  if  we  pass  a  plane  containing  this  line  AB,  it  may  be 
revolved  about  AB  until  it  contains  the  given  point. 

492.  THEOREM.     A  straight  line  and  an  external  point  determine 
a  plane.     [See  491,  486.] 


BOOK   VI 


253 


493.  THEOREM.     Three  points  not  in  a  straight  line,  determine  a 
plane. 

Because  two  of  the  points  may  be  joined  by  a  line;  then  this 
line  and  the  third  point  determine  a  plane.     [See  492.] 

494.  THEOREM.     Two  parallel  lines  determine  a  plane.      [See  91 
and  492.] 

495.  THEOREM.     Two  intersecting  straight  lines  determine  a  plane. 

Because  one  of  these  lines  and  a  point  in  the  second  line  de- 
termine a  plane  (492). 

And  this  plane  contains  the  second  line  (487). 


496.  THEOREM.  If  two  planes  are  parallel,  no  line  in  the  one  can 
meet  any  line  in  the  other.  [Def.  484.] 

NOTE.  A  plane  is  represented  to  the  eye  by  a  quadrilateral.  In 
some  positions  it  appears  to  be  a  parallelogram,  and  in  others,  a  trape- 
zoid.  The  eye,  however,  must  be  aided  by  the  imagination  in  really 
understanding  the  diagrams  of  Solid  Geometry.  Thus,  in  the  adjoining 
figure,  the  line  CN  is  perpendicular  to  the  plane 
FR,  and  it  is  perpendicular  to  every  line  in  FR 
drawn  through  N.  Consider  several  lines  drawn 
through  a  point  on  the  floor,  and  a  cane,  CN", 
occupying  a  vertical  position,  so  that  it  is  per- 
pendicular to  all  of  these  lines.  Then  every 
angle  CNX  is  a  right  angle,  though  to  the 
unskilled  eye  they  do  not  all  appear  to  be  right 
angles  in  the  diagram.  The  object  of  all  geo- 
metrical diagrams  is  that  the  eye  may  assist 
the  mind  in  grasping  truths  or  in  developing 

logical  demonstrations,  and  the  student  should  thoroughly  examine  every 
figure  until  he  completely  understands  the  relative  positions  of  its  parts, 
and  thus  trains  his  eye  to  see  three  dimensions  represented  in  a  plane. 
Photography  accomplishes  this,  and  we  should  be  as  familiar  with  the 
significance  of  a  geometrical  diagram,  as  with  a  picture. 

When,  during  the  process  of  a  demonstration  or  elsewhere,  it  becomes 
necessary  to  employ  a  plane  not  already  indicated,  it  is  customary  to  pass 
such  a  plane,  or  to  conceive  it  constructed. 


254  SOLID  GEOMETRY 

THEOREMS  AND  DEMONSTRATIONS 
POINTS.     LINES.     PLANES 

497.  THEOREM.     If  two  planes  intersect,  their  intersection  is  a 
straight  line, 

Given :  Intersecting  planes 
MN  and  BS. 

To  Prove :  Their  intersec- 
tion is  a  straight  line. 

Proof:  Suppose  A  and  B 
are  two  points  common  to 
both  planes.  Draw  AB,  a 
straight  line.  AB  is  in  plane 
ES  (?)  (487).  AB  is  in  plane 
MN  (?).  That  is,  AB  is  common  to  both  planes. 

Now,  if  there  were  a  point  outside  of  AB,  in  both  planes, 
these  planes  would  coincide  (?)  (492). 

That  is,  AB  contains  all  points  common   to   planes   MN 
and  ES. 

Hence,  AB  is  the  intersection  (482). 

That  is,  the  intersection  is  a  straight  line.  Q.E.D. 

498.  THEOREM.     If  two  straight  lines  are  parallel,  a  plane  contain- 
ing one,  and  only  one,  is  parallel 

to  the  other  line. 


Given  :  II  lines  AB  and  CD ; 
plane  MN  containing  CD. 

To  Prove:  plane  MN  II  to 
line  AB. 

Proof :  AB  and  CD  are  in 
the  same  plane  AD  (?)  (91). 
Plane  AD  intersects  plane 
MN  in  CD  (?)  (497). 


BOOK  VI 


255 


If  AB  ever  meets  MN,  it  must  meet  MN  in  CD ;  but  AB  can 
never  meet  CD  (Hyp.)- 

.-.  AB  can  never  meet  MN,  and  AB  is  II  to  MN  (?)  (484). 

Q.E.D. 

499.  THEOREM.     If  a  straight   line   is  parallel  to   a  plane,  and 
another  plane  containing  this  line  intersects  the  given  plane,  the  inter- 
section is  parallel  to  the  given  line. 

Given :  AB  II  to  MN  ;  plane 
AD  containing  AB  and  inter- 
secting plane  MN  in  CD. 
.  To  Prove :  AB  II  to  CD. 

Proof:  AB  and  CD  are  in 
the  same  plane  AD  (My p.). 
If  AB  ever  meets  CD,  it  must 
meet  CD  in  plane  MN;  but  AB  can  never  meet  MN  (Hyp.). 

/.  AB  can  never   meet  CD,  and  AB  is  II  to  CD  (?)  (91). 

Q.E.D. 

500.  THEOREM.     The  intersec- 
tions of  two  parallel  planes  by  a 
third  plane  are  parallel  lines. 

Given :  H  planes  AB  and 
CD  cut  by  plane  US  in  lines 
LM  and  PQ. 

To  Prove :  LM  II  to  PQ. 

Proof :  LM  and  PQ  are  in 
the  same  plane  B8  (Hyp.). 
Also,  LM  and  PQ  can  never 
meet  (?)  (496). 

.-.  LM  is  II  to  PQ  (?)  (91). 

Q.E.D. 


Ex.  1.  Why  are  the  usual  folds  in  a  sheet  of  paper  straight  lines? 
Ex.  2.   If  a  rod  is  held  parallel  to  the  pavement,  why  is  the  shadow 
parallel  to  the  rod  ? 


256 


SOLID   GEOMETRY 


M 


501.  THEOREM.     A  straight  line   perpendicular   to   each  of  two 
straight  lines  at  their  intersection  is  perpendicular  to  the  plane  of  the 
lines. 

Given :  AF  _L  to  BF  and  CF 
at  F\  plane  MN  containing  BF 
and  CF. 

To  Prove :  AF  _L  to  plane  MN. 

Proof  :  In  plane  MN  draw  BC-, 
draw  also  DF  from  F  to  any 
point,  D,  in  BC. 

Prolong  AFto  X,  making  FX 
=  AF,  and  draw  AB,  AD,  AC, 
BX,  DX,  CX. 

Now,  BF  and  CF  are  _L  to  AX  at  its  midpoint  (Hyp.  and 
Const.). 

In   A  ABC  and   BCX,   AB  =  BX,  AC  =  CX  (?)   (67),  and 

BC=BC  (?).    .-.  A  ABC  =  A  BCX  (?)  (58). 

Also,    in  A  ABD  and   BDX,  /.  ABC  =  /.  CBX  (?)    (27), 
BD  =  £D  (?),  and  AB  =  5JT  (?). 

.-.  A  ABD  =  A  BDX  (?)  (52)  and  ^1Z>  =  DX  (?). 
.-.  D^  is  J_  to  ^X  (?)  (70). 
That  is,  AF  is  J_  to  all  lines  in  MN  through  F. 
Consequently,  AF  is  J.  to  plane  MN  (?)  (483).  Q.E.D. 

502.  THEOREM.     All  straight  lines  perpendicular  to  a  line  at  one 
point  are  in  one  plane,  which  is  per- 
pendicular to  this  line  at  this  point. 

Given :  AB  _L  to  BC,  BD,  BE,  etc. ; 
plane  MN  containing  BC  and  BD. 

To  Prove:  BE  is  in  the  plane 
MN  and  MN  is  ±  to  AB  at  B. 

Proof  :  Pass  plane  AE  contain- 
ing AB  and  BE,  and  intersecting 
MN  in  line  BX.  L-=- 1N 

Now,  AB  is  J_  to  MN  (?)  (501). 


BOOK   VI 


257 


That  is,  plane  MN  is  J-  to  AB. 
.-.  AB  is  J_  to  BX  (?)  (489). 

But  AB  is  _L  to  BE  (Hyp.).     That  is,  BX  and  BE  are  both 
-i.  to  AB,  in  the  plane  AE,  at  B. 

Hence,  BE  and  BX  coincide  (?)  (43). 

That  is,  BE  is  in  the  plane  MN.  Q.E.D. 

503.  THEOREM.     Through  a  point  in  a  straight  line  one  plane  can 
be  passed  perpendicular  to  the  line,  and  only  one.     (502.) 

504.  THEOREM.     Through  an   external  point   one  plane   can   be 
passed  perpendicular  to  a  given  straight  line, 

and  only  one. 

Given :  The  line  AB  and  point  P  out- 
side of  AB. 

To  Prove :  Through  P,  one  plane  can 
be  passed  JL  to  AB,  and  only  one. 

Proof :   I.    Draw  from  P,  PC  _L  to  AB, 
and  at  C  draw  CX,  another  line  _L  to  AB. 
PC  and  CX  determine  a  plane  MN  (?). 
Plane  MN  contains  P  and  is  _L  to  AB  (?)  (501). 

II.    Only  one  line  -L  to  AB  can  be  drawn  from  P  (?)  (71). 

And  only  one  plane  J_  to  AB  can  be  passed  at  C  (?)  (503). 

That   is,    MN  is   the  only  plane   J-   to  AB,  that   can   be 

passed  through  P.  Q.E.D. 


N 


IB 


Ex.  1.  If  two  lines  are  each  parallel  to  a  plane,  are  the  lines  neces- 
sarily parallel  ? 

Ex.  2.  If  two  planes  intersect,  how  can  a  line  be  drawn  through  a 
given  point  that  shall  be  parallel  to  both  planes  ? 

Ex.  3.  How  many  planes  are  determined  by  two  intersecting  lines 
(a  and  6)  and  two  points  (R  and  S)  not  in  the  plane  of  the  lines? 


258 


SOLID   GEOMETRY 


B 


505.  THEOREM.     Two  planes  perpendicular  to  the  same  straight 
line  are  parallel. 

Given  :  Planes  MN  and  OP  1_  to  AB. 
To  Prove :  MN  II  to  OP. 

Proof :  If  the  planes  MN  and  OP  are 
not  II,  they  will  meet  when  sufficiently 
extended  (?)  (Def.  484). 

Then  there  would  be  two  planes  from 
the  same  point  _L  to  AB^JL  by  hyp.).  / / 

But  this  is  impossible  (?)  (504).  ~~]~~  P 

.-.  the  planes  never  meet  and  therefore 
they  are  parallel  (Def.  484).  Q.E.D. 

506.  THEOREM.     At  a  given  point  in  a  plane  one  line  can  be  drawn 
perpendicular  to  the  given  plane,  and  only  one. 

Given:  Plane  MN  and  point  P 
within  it. 

To  Prove :  One  line  can  be 
drawn  _L  to  MN  at  P,  and  only  one. 

Proof :  I.  In  MN  draw  any  line 
AB,  through  P. 

Suppose  plane  CD  be  passed  J_ 
to  AB  at  P,  meeting  the  plane  MN 
in  CE.  In  plane  CD  draw  PR  _L  to 
CE,  from  P.  Now,  AB  is  _L  to  plane  CD  (Const.). 

.-.  AB  is  -L  to  PR  (?)  (489). 

PR  is  _L  to  CE  (Const.).     .-.  PR  is  _L  to  plane  MN  (?)  (501). 

II.  Suppose  another  line  PX  to  be  _L  to  plane  MN  at  P. 
Then  PX  and  PR  would  determine  a  plane  CD,  intersecting 
plane  MN  in  CE  (?)  (495  and  497). 

PR  and  PX  would  then  both  be  J_  to  CE  at  P  (?)  (489). 

But  this  is  impossible  (?)  (43). 

That  is,  PR  and  PX  coincide,  and  PR  is  the  only  line  _L 
to  plane  MN  at  P.  Q.E.D. 


BOOK  vi 


259 


—.1 


507.   THEOREM.     Through  a  given  external  point  one  line  can  be 
drawn  perpendicular  to  a  given  plane,  and  only  one. 

Given:  Plane  MN  and  point  P 
outside  of  it. 

To  Prove :  One  line  can  be  drawn 
through  P  _L  to  MN,  and  only  one. 

Proof  :  I.  In  plane  MN  draw  any 
line  AB.  Suppose  a  plane  GH  be 
passed  through  P  J_  to  AB,  meet- 
ing plane  'MN  in  EC,  and  AB  at  C. 
In  plane  GH  draw  PR  _L  to  KG  and 
prolong  PR  to  X,  making  RX=PR. 
Draw  RD  to  any  point  in  AB, 
except  C.  Draw  PC,  PD,  CX,  DX. 
Now  RC  is  -L  to  PX  at  its  midpoint  (Const.). 

Also  AB  is  _L  to  plane  GH  (Const.). 
Hence,  A  DCP  and  DCX  are  rt.  A  (?)  (489). 
In  rt.  A  DCP  and  DCX,  DC  =  DC  (?),  PC=  CX  (?)  (67). 
.-.  A  DCP  =  A  DCX  (?)  (53).     Hence,  PD  =  XD  (?). 
.-.  RD  is  _L  to  PX  (?)  (70).     That  is,  PR  is  J_  to  RC  and 
RD,  in  plane  MN. 

Consequently,  PR  is  _L  to  plane  MN  from  P  (?)  (501). 

II.    Suppose  there  is  another  line  PL  _L  to  plane  MN  from 
p.     Then  PR  and  PL  determine  a  plane  GH  (?). 

This  plane  intersects  plane  MN  in  EC  (?). 
PR  and  PL  would  then  both  be  J_  to  EC  (?)  (489). 

But  this  is  impossible  (?)  (71). 

That  is,  PR  and  PL  coincide  and  therefore  PR  is  the  only 
line  J.  to  plane  MN  from  P.  Q.E.D. 


Ex.  In  the  figure  of  507  name  the  six  right  angles  at  C. 


260 


SOLID   GEOMETRY 


508.  THEOREM.     If  a  plane  is  perpendicular  to  a  line  in  another 
plane,  any  line  in  the  first  plane  perpendicular  to  the  intersection  of 
the  planes  is  perpendicular  to  the  second  plane. 

Proof:  Identical  with  the  proof  of  507,  I. 

509.  THEOREM.     If  a  plane  is  perpendicular  to  one  of  two  parallel 
lines,  it  is  perpendicular  to  the  other  also. 

Given:  Plane  MN  JL  to  line  AB,  and 
AB  II  to  CP. 

To  Prove  :  CP  J_  to  plane  MN. 

Proof :  Lines  AB  and  CP  determine  a 
plane  (?)  (494). 

Pass     this     plane     BC,    intersecting 
plane  MN  in  line  BP. 

Draw  BX  _L  to  BP,  in  plane  MN. 

AB  is  _L  to  BX  (?)  (489).    .-.  BX  is  J_  to  plane  BC  (?)  (501). 

BP  is  _L  to  AB  (?)  (489).     .-.  BP  is  _L  to  CP  (?)  (95). 

That  is,  plane  BC  is  J_  to  BX  and  CP  is  J_  to  the  intersec- 
tion BP. 

.'.  CP  is  _L  to  plane  MN  (?)  (508).  Q.E.D. 

510.  THEOREM.     Two  lines  perpendicular  to  the  same  plane  are 
parallel. 

Given  :  Lines  AB  and  CD  J_  to  plane 

MN. 

To  Prove :  AB  II  to  CD. 

Proof:  Through  D,  the  foot  of  CD, 
draw  DX  II  to  AB. 

Then  DX  is  _L  to  plane  MN  (?)  (509).  ~*N 

But  CD  is  J_  to  plane  MN  at  D  (  Hyp. ) . 

/.  DX  and  DC  coincide  (?)  (506).     That  is,  AB  is  II  to  CD. 

Q.E.D. 


M 


:X 


BOOK  VI 


261 


511.   THEOREM.     Two  straight  lines  that  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 

Given :  Lines  CD  and  EF  each  II  to 
AB. 

To  Prove :  CD  II  to  EF. 


Proof :  Suppose  plane  MN  be  passed 
to  AB. 

.-.  MNis  _LtoCZ>andto^F(?)  (509). 
.-.  CD  is  II  to  EF  (?)  (510).        Q.E.D. 


M. 


B 


M 


:E 


N 


512.  THEOREM.     A  line  perpendicular  to  one  of  two  parallel  planes 
is  perpendicular  to  the  other  also. 

Given :  Plane  MN  II  to  plane 
E8\  AP  _L  to  plane  ES. 

To  Prove :  AP  _L  to  MN. 

Proof :  Through  AP  pass 
any  two  planes,  AB  and  AC, 
intersecting  MN  in  AD  and 
AE,  and  intersecting  US  in  PB 
and  PC. 

AD  is  II  to  PB,  and  AE  is  II  to 
PC(?)  (500). 

AP  is  _L  to  PB  and  PC  (?)  (489). 

/.  ^4P  is  _L  to  AD  and  AE  (?)  (95). 

.-.  ^iP  is  _L  to  plane  JOT  (?)  (501).  Q.E.D. 

513.  THEOREM.     If  two  planes  are  each  parallel  to  a  third  plane, 
they  are  parallel  to  each  other. 

Proof :  Draw  a  line  _L  to  the  third  plane.     This  line  is  _L  to 
each  of  the  other  planes  (?).      .-.  the  planes  are  II  (?)  (505). 


a 


Ex.  Why  are  not  AB,  CD,  EFiu  figure  of  511  all  parallel? 


262 


SOLID   GEOMETRY 


514.  THEOREM.     If  two  intersecting  lines  are  each  parallel  to  a 
plane,  the  plane  of  these  lines  is  parallel  to  the  given  plane. 

Given :  Intersecting  lines  AS 
and  AC  in  plane  JOT;  each  line 
II  to  plane  PQ. 

To  Prove:  Plane  MN  II  to 
plane  PQ. 

Proof :  Draw  AE  JL  to  MN  at 
A,  meeting  PQ  at  R.  Through 
AE  and  AB  pass  plane  AS,  and 
through  AE  and  AC  pass  plane 
AT,  intersecting  plane  PQ  in 
E8  and  ET,  respectively. 

AB  is  II  to  ES,  and  AC  is  II  to  ET  (?)  (499). 

But,  AE  is  _L  to  AB  and  AC  (?)  (489). 

/.  AE  is  -L  to  ES  and  ET  (?)  (95). 

Hence,  AE  is  _L  to  plane  PQ  (?)  (501). 

.-.  plane  MN  is  II  to  plane  PQ  (?)  (505).  Q.E.D. 

515.  THEOREM.     If  two  angles,  not  in  the  same  plane,  have  theii 
sides  parallel  each  to  each,  and  ex- 
tending in  the  same  directions  from 

their  vertices,  the  angles  are  equal 
and  the  planes  are  parallel. 

Given :  Z  BAG  in  plane  MN 
and  Z  EDF  in  plane  PQ  ;  AB  II 
to  DE;  AC  II  to  DF,  and  extend- 
ing in  the  same  directions. 

To  Prove: 

I.    Z  BAC=  Z  EDF. 
II.    Plane  MN  II  to  plane  PQ. 

Proof :  I.     Take  DE  and  AB  equal,  and  DF  and  AC  equal, 

Draw  AD,  BE,  OF,  BC,  EF. 
The  figure  ABED  is  a  O   (?)  (135).      .'.    AD  =  BE  (?), 


BOOK  VI  263 

Also  ACFD  is  a  O  (?).      .'.  AD  =  CF  (?). 
.'.  BE=  CF  (?).     Also  AD  is  II  to  BE  and  AD  is  II  to  CF  (?). 
.-.  BE  is  II  to  CF  (?)  (511).      /.  BCFE  is  a  O  (?)  (135). 

Now,  in  A  ABC  and  DEF,  AB  =  DE  (?)  ;  AC  =  DF  (?),  and 
BC  =  ^F  (?)  (130).   /.  A  ABC  =  A  DEF  (?)  (58). 


II.  AB  is  II  to  plane  PQ  and  AC  is  II  to  plane  PQ  (?)  (498). 
.-.  plane  MN  is  II  to  plane  PQ  (?)  (514).      Q.E.D. 

516.   THEOREM.    If  three  parallel  planes  intersect  two  straight 
lines,  the  corresponding  intercepts  are  proportional. 

Given  :  Parallel  planes,  LM,  NP, 
QR,  intersecting  line  AB  at  A,  E,  B, 
and  CD  at  C,  F,  D,  respectively. 

To  Prove  :   AE  :  EB  =  CF :  FD.  | f_ 

Proof:  Draw  .BC,  meeting  plane      /   U-"""G  ' 
^P  at  G. 

Through  AB  and  BC  pass  a 
plane  cutting  LM  in  AC  and  NP 
in  EG.  Through  BC  and  CD  pass 
a  plane  cutting  NP  in  GF  and  QR 
in  BD.  Now,  EG  is  II  to  AC  and  GF  is  II  to  BD  (?)  (500), 


Consequently,  AE  :  EB=  CF  :  FD  (Ax.  1).  Q.E.D 

Also,  AB  and  CD  are  intercepts  between  planes  LM  and  QR 


And  ^  =  (™\  =  —  (?)  (305). 
AB       \CBj        CD 


Ex.     Prove,  in  figure  of  516,  AB:  CD  =  AE  :  CF  =  EB:  FD. 


264 


SOLID   GEOMETRY 


R  J 


517.  THEOREM.     The  projection  of  a  straight  line  upon  a  plane  is 
a  straight  line.* 

Given:  Line  AB  and  plane 

air. 

To  Prove  :  The    projection 
of  AB  on  MN  is  a  straight  line. 

Proof :  Draw  PJ  JL  to  plane 
MN  from  any  point  P,  in  AB. 

AB    and    PJ    determine    a 
plane  (?)  (495).  

This    plane    AD    intersects 
plane  MN  in  a  straight  line  CD  (?). 

Now  in  plane  AD  draw  XR  I!  to  PJ  from   JT,  any  other 
point  in  AB.     XR  is  -L  to  MN  (?)  (509). 

Now  R  is  the  projection  of  X  (?)  (485). 
.-.  CD  is  the  projection  of  AB  (?)  (485). 

That  is,  the  projection  of  AB   upon   the  plane  MN  is  a 
straight  line.  Q.E.D. 

518.  COR.     A  line  and  its  projection  upon  a  plane  are  in  the  same 
plane. 

519.  THEOREM.     A  line  not  parallel  to  a  plane  is  longer  than  its 
projection  upon  the  plane. 

Given:  A  plane  and  line  LN  not  II  to 
the  plane,  and  DE  the  projection  of  LN 
upon  the  plane. 

To  Prove:  LN>DE.  Proof:  Drawxz) 
and  NE.  Draw  LX  _L  to  NE  from  i,  in 
the  plane  LE.  LD  and  NE  are  -L  to  the 
plane  (Def.  of  projection,  485). 

LXED  is  a  rectangle  (?)  (166). 
Now  LN  >  LX  (?)  (77).     But  LX  =  DE  (?)  (130). 

Hence,  LN  >  DE  (Ax.  6).  Q.E.D. 


*  Except  only  if  the  given  straight  line  is  a  normal  to  the  given  plane. 


BOOK   VI 


265 


520.   THEOREM.    Of  all  lines  that  can  be  drawn  to  a  plane  from 
a  point  : 

I.   The  perpendicular  is  the  shortest. 
II.   Oblique  lines  having  equal  projections  are  equal. 
Ill  ,  Equal  oblique  lines  have  equal  projections. 

IV.   Oblique  lines  having  unequal  projections  are  unequal,  and  the 
line  having  the  greater  projection  is  the  longer. 

V.   Unequal  oblique  lines  have  unequal  projections,  and  the  longer 
line  has  the  greater  projection. 

I.  Given:  Plane  MN;  point 
P  ;  PR  _L  to  MN  ;  any  other  line 
from  P  to  plane  MN,  as  PA. 

To  Prove  :  PR  <  PA. 

Proof:    Draw  AR.     Now  PR 

(?)  ;  PA  is  not  A.  to  AR 


N 


II.  Given:    Oblique  lines  PA 

and  PB,  whose  projections  A  R  and  BR  are  equal. 

To  Prove  :  PA  =  PB. 

Proof  :  The  rt.  A  PEA  and  PRB  are  =.     (Explain.) 

III.  Given  :  Equal  oblique  lines  PA  and  PB. 

To  Prove  :  Their  projections,  AR  and  BR,  are  equal. 
Proof:  The  rt.  A  PRA  and  PRB  are  =  .     (Explain.) 

IV.  Given  :  Oblique  lines  PC  and  PA\  proj.  EC  >  proj.  RA 
To  Prove  :  PC  >  PA. 

Proof:  In  A  PCR,  take  on  RC,  RX  =  RA,  and  drawPJT. 
Now  PC  >  PX  (?)  (76,  III).     But  PA  =  PX  (?)  (520,  II) 

.-.  PC>  PA  (Ax.  6). 

Y.    Given  :  Unequal  oblique  lines,  PC  >  PA. 
To  Prove  :  Projection  EC  >  projection  RA. 
Proof  :  By  reductio  ad  absurdum.     (See  88,  89). 


266 


SOLID   GEOMETRY 


521.  THEOREM.  The  acute  angle  that  a  line  makes  with  its  own 
projection  upon  a  plane  is  the  least  angle  that  the  line  makes  with 
any  line  of  the  plane. 

Given  :  AB,  any  line  meeting  plane 
MN  at  B  ;  BP,  its  projection  upon  MN ; 
BD,  any  other  line  in  MN,  through  B. 

To  Prove :  Z  ABP  <  Z  ABD. 

Proof:  On  BD  take  BX=BP  and 
draw  AX.  In  A  ^4PJ5  and  ABX,  AB  = 
AB  (?)  ;  BP  =  BX  (?)  (Const.). 

But  AP  <  AX  (?)  (520,  I). 


.'.  Z  ^IBP  <  Z  ^BD  (?)  (87).  Q.E.D. 

522.   THEOREM.   Through  a  given  point  one  plane  can  be  passed 
parallel  to  any  two  given  lines  in  space,  and  only  one. 

Given :   Point  P;  two  lines,  AB  and  CD. 
To  Prove :    Through  P  one  plane  can 


be  passed  II  to  AB  and  CD,  and  only 
one. 

Proof  :  I.    Through  P  draw  a  line  II  to 
AB  and  another  II  to  CD.     Pass  a  plane 
,  containing  these  lines. 


MN  is  II  to  both  AB  and  CD  (?)  (498). 

II.    Only  one  line  can  be  drawn  through  P  II  to  AB,  and 
only  one,  II  to  CD  (?)  (92). 

.-.  there  is  only  one  plane  (?)  (495).  Q.E.D. 


Ex.  1.  If  a  line  meets  a  plane,  with  what  line  in  the  plane  does  this 
line  make  the  greatest  angle  ? 

Ex.  2.  In  the  figure  of  515,  prove  A  D  parallel  to  the  plane  CE. 
Also  BE  parallel  to  the  plane  A  F. 

Ex.  3.   In  the  figure  of  516,  if  AE  =  Q,EB  =  5,  CD  =  16J,  find  DF. 


BOOK  VI  267 

523.  THEOREM.   Through  a  given  point  one  plane  can  be  passed 
parallel  to  a  given  plane,  and  only  one.  x 

Given:  (?).     To  Prove:  (?).  l~     ~        V 

Proof :  I.   Suppose  PR  be  drawn  J_  to      /  i  \y 

AB-,  and  XY  be  passed  _L  to  PR  at  P. 

Then  XY  is  II  to  AB  (?)  (505). 

II.    Only  one  line  _L  to  AB  can  be        ' 
drawn  from  P    (?)  (507).     Only  one 
plane  _L  to  PR  can  be  passed  at  P  (503). 

/.  only  one  plane  can  contain  P  and  be  II  to  AB.          Q.E.D. 

524.  THEOREM.     Parallel   lines    included 
between   parallel  planes  are  equal. 

Given  :  (?).     To  Prove:  (?). 
Proof  :  AB  and  CD  determine  a  plane 
(?).     This  plane  intersects  RS  and  PQ  in 
lines  AC  and  BD,  which  are  II  (?)  (500). 
.'.ABDC    is    a  O     (?). 
Hence,  AB  =  CD  (?).  Q.E.D.  v 

525.  THEOREM.    The  plane  perpendicular  to  a  line  at  its  midpoint 
is  the  locus  of  points  in  space,  equally  distant  from  the  extremities 
of  the  line. 

Given  :   RS  _L  to  AB  at  its  midpoint,  M. 

To  Prove:    (?). 

Proof  :  (1)  Take  P,  any  point  in  RS. 
Draw  PM,  PA,  PB.  PM  is  J_  to  AB  (?) 
(489).  .-.  PA  =  PB  (?)  (67). 

That  is,  any  point  in  RS  is  equally 
distant  from  A  and  B. 

(2)    Take  P;,  any  point  outside  of  RS.     Draw  PfM. 

P!M  is  not  _L  to  AB  (?)  (502).     .'.  P',  any  point  outside  of 
plane  RS,  is  not  equally  distant  from  A  and  B  (?)  (68). 

Hence,  plane  RS  is  the  locus  of  points  in  space  equally  dis- 
tant from  A  and  B  (?)  (179).  Q.E.D. 


M 


/•P- 


268 


SOLID   GEOMETRY 


526.  THEOREM.     The   locus  of   points  in   space  equally  distant 
from  all  the  points  in  the  circumference  of  a  circle  is  the  line  perpen- 
dicular to  the  plane  of  the  circle  at  its  center. 

Given  :  (?).     To  Prove  :  (?). 

Proof  :  I.  Any  point  in  AC  is  equally 
distant  from  all  the  points  in  the  cir- 
cumference of  the  circle  (?)  (520,  II). 

II.  Any  point  equally  distant  from 
all  points  of  the  circumference  of  the 
circle  is  in  AC  (?)  (520,  III).  M 

/.  AC  is  the  required  locus  (179).  Q.E.D. 

Ex.  1.  What  is  the  locus  of  points  equally  distant  from  two  given 
points  ? 

Ex.  2.  What  is  the  locus  of  points  equally  distant  from  three  given 
points  ? 

527.  The  distance  from  a  point  to  a  plane  is  the  length  of 
the  perpendicular  from  the  point  to  the  plane. 

Thus,  the  word  "distance,"  referring  to  the  shortest  line  from  a  point 
to  a  plane,  implies  the  perpendicular. 

The  inclination  of  a  line  to  a  plane  is  the  angle  between 
the  line  and  its  projection  upon  the  plane. 


ORIGINAL  EXERCISES 

1.  Through  one  straight  line  a  plane  can  be  passed  parallel  to  any 
other  straight  line  in  space,  and  only  one. 

Through  a  point  of  the  first  line  draw  a  line  ||  to  the  second. 

2.  Two  parallel  planes  are  everywhere  equally  distant. 

3.  If  a  line  and   a  plane    are  parallel,  another  line  parallel  to  the 
given  line  and  through  any  point  in  the  given  plane  lies  wholly  in  the 
given  plane. 

Through  the    given  line  and  the  point  P  pass  a  plane  cutting  the 
given  plane  in  PX.     Use  499. 


BOOK  VI  269 

4.  A  straight  line  parallel  to  the  intersection  of  two  planes,  but  in 
neither,  is  parallel  to  both  planes. 

5.  If  two  straight  lines  are  parallel   and  two   intersecting  planes 
are  passed,  each  containing  one  of  the  lines,  the  intersection  of  these 
planes  is  parallel  to  each  of  the  given  lines. 

6.  If  three  straight  lines  through  a  point  meet  the  same  straight  line, 
these  four  lines  all  lie  in  the  same  plane. 

y  7.   If  a  straight  line  meets  two  parallel  planes,  its  inclinations  to  the 
planes  are  equal. 

^8.   Two  parallel  planes  can  be  passed,  each  containing  one  of  two 
given  lines  in  space.     Is  this  ever  impossible? 

9.   If  each  of  three  straight  lines  intersects  the  other  two,  the  three 
lines  all  lie  in  a  plane. 

10.  The  projections  of  two  parallel  lines  on  a          ^rx^fv 
plane  are  parallel.  |      JD 

Proof:  AB  is  ||  to  CD  (?).      AE  is  ||  to  CG  (?).      /ELjp       H\ 
.-.  planes  AF  and  CH  are  ||  (?)  ;  etc. 

11.  If  two  lines  in  space  are  equal  and  parallel,  their  projections  on 
a  plane  are  equal  and  parallel. 

12.  If  a  plane  is  parallel  to  one  of  two  parallel  lines,  it  is  parallel  to 
the  other. 

13.  If   a   straight  line  and  a  plane  are  perpendicular  to  the  same 
straight  line,  they  are  parallel. 

14.  Equal  oblique  lines  drawn  to  a  plane  from  one  point  have  equal 
inclinations  with  the  plane. 

15.  If  a  line  and  a  plane  are  both  parallel  to  the  same  line,  they  are 
parallel  to  each  other. 

16.  Four  points  in  space,  A,  B,   C,  A  are  joined,  and  these  four 
lines  are  bisected.     Prove  that  the  four  lines  joining 

(in  order)  the  four  midpoints  of  the  first  lines  form 
a  parallelogram. 

Proof:  Pass  plane  DP  through  points  A,  D,  B, 
and  plane  DX  through  points  B,  C,  D,  —  these 
planes  intersecting  in  BD.  ST  is  ||  to  BD  and  = 
I  BD  (?) ;  etc. 


270  SOLID   GEOMETRY 

17.  If  a  plane  is  passed  containing  a  diagonal  of  a  parallelogram 
and  perpendiculars  be  drawn  to  the  plane  from  the 

other  vertices  of  the  parallelogram,  they  are  equal. 

To  Prove :  A  E  =  CF.  Proof :  Draw  diagonal 
AC.  Draw  EO  and  OF  in  plane  MN.  EO,  OF, 
and  EOF  are  projections;  etc. 

18.  If  from  the  foot  of  a  perpendicular  to  a 
plane,  a  line  be  drawn  at  right  angles  to  any  line 
in  the  plane,  the  line  connecting  this  point  of  inter- 
section with  any  point  in  the  perpendicular  is  per- 
pendicular to  the  line  in  the  plane. 

Given:  AB  J_  to  plane  RS;  EC  JL  to  DE  in  the 
plane;  PC  drawn  from  C  to  P,  in  AB. 

To  Prove:   PC  is  JL  to  DE. 

Proof :   Take  CD  =  CE,  draw  PD,  PE,  BD,  BE.     EC  is  JL  to  DE  at 
its  midpoint  (?).     .-.  BD  =  BE  (?).     PD  =  PE  (?)  (520,  II). 

.-.  PC  is  J_  to  DE  (?)  (70). 

19.  A  line  PB  is  perpendicular  to  a  plane  at  B,  and  a  line  is  drawn 
from  B  meeting  any  line  DE,  of  the  plane,  at  C.     If  PC  is  perpen- 
dicular to  DE,  BC  is  perpendicular  to  DE. 

20.  Are   two   planes  that   are    parallel   to    the   same    straight  line 
necessarily  parallel  ? 

21.  If  each  of  two  parallel  lines  is  parallel  to  a  plane,  is  the  plane  of 
these  lines  also  parallel  to  the  given  plane? 

22.  Is  a  three-legged  chair  always  stable  on  the  floor?    Why?    Is  a 
four-legged  chair  always  stable  ?    Why  ? 


23.  What  is  the  locus  in  space  of  points  equally  distant  from  two 
parallel  planes  ?     From  two  parallel  lines  ? 

24.  What  is  the  locus  of  points  in  space  at  a  given  distance  from  a 
given  plane  ? 

25.  What  is  the  locus  of  points  in  a  plane  at  a  given  distance  from 
an  external  point? 

26.  What  is  the  locus  of  points  in  space  equally  distant  from  two 
points  and  equally  distant  from  two  parallel  planes? 


BOOK  VI  271 


27.  What  is  the  locus  of  points  ir.  space,  equally  distant  from  the 
vertices  of  a  given  triangle  ? 

28.  What  is  the  locus  of  all  straight  lines  perpendicular  to  a  given 
straight  line  at  a  given  point? 

29.  What  is  the  locus  of  all   lines  parallel  to  a  given  plane  and 
drawn  through  a  given  point? 

30.  If  the  points  in  a  line  satisfy  one  condition  and  the  points  in  a 
plane  satisfy  another  condition,  what  will  be  true  of  their  intersection? 
What  will  be  true  if  they  do  not  intersect? 

31.  If  the  points  in  one  plane  satisfy  one  condition  and  the  points 
in  another  plane  satisfy  another  condition,  what  is  true  of  their  inter- 
section ?    What  is  true  if  the  planes  are  parallel  ? 


32.  To  construct  a  plane  perpendicular  to  a  given  line  at  a  given 
point  in  the  line. 

33.  To  construct  a  plane  perpendicular  to  a  given  line  through  a 
given  external  point. 

34.  To  construct  a  line  perpendicular  to  a  given  plane,  through  a 
given  point  in  the  plane. 

35.  To  construct  a  line  perpendicular  to  a  given  plane,  through  a 
given  external  point. 

36.  To  construct  a  plane  parallel  to  a  given  plane,  through  a  given 
point. 

37.  To  construct  a  number  of  equal  oblique  lines  to  a  plane  from  a 
given  external  point. 

38.  To  construct  a  line  through  a  given  point  parallel  to  a  given 
plane. 

39.  To  construct  a  line  through  a  given  point  and  parallel  to  each  of 
two  given  intersecting  planes. 

40.  To  construct  a  plane  containing  one  given  line  and  parallel  to 
another. 

41.  To  construct  a  plane  through  a  given  point  parallel  to  any  two 
given  lines  in  space. 


272  SOLID   GEOMETRY 

42.  To  construct  a  line  through  a  given  point  in  space  which  will 
intersect  two  given  lines  not  in  the  same  plane. 

When  is  there  no  such  line?     Is  there  ever  more  than  one? 

43.  To  find  a  point  in  a  plane  such  that  the  sum  of  the  two  lines 
joining  it  to  two  fixed  points  on  one  side  of  the  plane 

shall  be  the  least  possible.  ,g 


:      rw  o        ne  - 

long  it  to  X,  making   CX  =  AC.     Draw  BX,  meet-  r  j\ 

ing  plane  at  P.     Draw  A  P.      Take  any  other  point  /£•     v£, 

R  in  plane  MN.  J 

Statement:  AP  +  PB<AR  +  RB.    Etc.  x*' 


44.  To  find  a  point  in  a  given  line  equally  distant  from  two  given 
points,     Is  this  ever  impossible  ? 

45.  To  find  a  point  in  a  given  plane  equally  distant  from  three  given 
points.     Is  this  ever  impossible? 

46.  To  find  the  line  whose  points  are  equally  distant  from  two  given 
points,  and  at  a  given  distance  from  a  given  plane. 

47.  To  find  the  point  equally  distant  from  two  given   points  and 
equally  distant  from  three  other  given  points;  or,  to  find  the  point  equally 
distant  from  the  ends  of  a  given  line,  and  also  equally  distant  from  the 
vertices  of  a  given  triangle. 

When  is  there  no  such  point? 

48.  To  find  the  one  point  equally  distant  from  four  given  points  not 
in  the  same  plane. 


Given  :   The  four  points,  A,  B,  C,  D. 

Required :   To  find  a  point  equally  distant  from  all 
of  them.  /    [A M 

/p./  y:  / 

Construction:    Pass  plane  CM,  containing^,  B,  C,         {of / 

and  plane  CN,  containing  A,   Z),    C.     Find    O,  the         j  /    Q'    ^/ 

center  of  the  O  containing  A ,  B,  C.     Find  P,  similarly.      ^L i 

Draw  the  locus  of  points  equally  distant  from  A ,  B, 

C.     (Consult  526.)      Draw  the  locus  of  points  equally 

distant  from  A,  D,  C.     The  plane  _L  to  AC  at  its  mid-point  contains 

both  these  loci.     (Explain.)     Hence,  OX  and  PX  intersect  (?)     .-.  X  is 

the  point. 


BOOK  VI 


273 


DIHEDRAL   ANGLES 

528.  A  dihedral  angle  is  the  amount  of  divergence  of  two 
intersecting  planes.     The  edge  of  the  dihedral  angle  is  the 
line  of  intersection  of  the  planes.     The  faces  of  the  dihedral 
angle  are  the  planes. 

The  intersecting  planes  AG  and  ED  form  the 
dihedral  angle  whose  edge  is  EG.  This  dihedral  A 

angle  is  named  A-GE-D;  or,  when  there  is  only 
one  dihedral  angle  at  the  edge,  it  may  be  called 
"the  angle  .EG." 

529.  Adjacent   dihedral  angles  are  two 

dihedral  angles  that  have  the  same  edge 
and  a  common  face  between  them. 

Vertical  dihedral  angles  are  two  dihedral 
angles  that  have  the  same  edge,  and  the 
faces  of  one  are  the  extensions  of  the  faces 
of  the  other. 

530.  The  plane  angle  of  a  dihedral  angle 
is  the  angle  formed  by  two  straight  lines, 
one  in  each  face,  and  perpendicular  to  the 
edge  at  the  same  point. 

If  PM  is  in  plane  AG  and  perpendicular  to  EG, 
and  PN  is  in  plane  ED  and  perpendicular  to  EG  at 
P,  the  angle  MPN  is  the  plane  angle  of  the  dihe- 
dral angle  EG. 

531.  If  one  plane  meets  another,  making 
one  adjacent  dihedral  angles  equal,  these 
angles  are  right  dihedral  angles. 

One  plane  is  perpendicular  to  another  plane  if  the  dihedral 
angle  formed  by  the  two  planes  is  a  right  dihedral  angle. 

532.  Two  dihedral  angles  are  equal  if  they  can  be  made 
to  coincide. 


274 


SOLID  GEOMETRY 


A  dihedral  angle  is  acute,  right,  or  obtuse  according  as  its 
plane  angle  is  acute,  right,  or  obtuse. 

Dihedral  angles  are  complementary  or  supplementary 
according  as  their  plane  angles  are  complementary  or  supple- 
mentary. 

533.  THEOREM.     The   plane  angles   of   a 
dihedral  angle  are  all  equal. 

Given  :  Z  EFG,  the  plane  Z  of  dihedral 
Z  BC,  at  F,  and  Z  RST,  the  plane  Z  at  S. 

To  Prove:  Z  EFG  =  Z  RST. 

Proof:  EF  is  ||  to  US,  and  FG  is  II  to  8T 
(?)  (93). 

/.  /L  EFG  =  Z  RST    (?)    (515).     Q.E.D. 

534.  THEOREM.     The  plane   of  the   plane 
angle  of  a  dihedral  angle  is  perpendicular  to 
the  edge.     (See  501.) 

535.  THEOREM.     Two  dihedral  angles  are  equal  if  their  plane 
angles  are  equal. 

Given :  Dihedral  A  CB  and  cfBf  whose  plane  A  EBD  and 
E'B'D'  are  equal. 

To  Prove: 

Dih.  Z  CB  =  dih.  Z  CrBf. 

Proof:  CB  is  _L  to  plane  EBD, 
and  C'B'  is  _L  to  plane  E'B'D' 
(?)  (534). 

Apply  dih.  Z  CfBf  to  dih. 
Z  CB  so  that  the  plane  Z  E'B'D' 
coincides  with  its  equal  Z  EBD. 

C'B'  will  coincide  with  CB 
(?)  (506). 

.*.  plane  CfDr  will  coincide  with  plane  CD  and  plane  CfE] 
will  coincide  with  plane  CE  (?)  (495). 

Hence,  dih.  Z  CB  =  dih.  Z  C'B'  (?)  (532).         Q.E.D. 


D1 


BOOK  VI 


275 


536.  THEOREM.   If  two   dihedral  angles  are  equal,  their  plane 
angles  are  equal. 

Proof :  Superpose  dih.  Z  C'B'  upon  its  equal  dih.  Z  OB, 
making  B'  fall  on  B,  and  edge  B'C'  on  BC.  Then  face  CfDr 
will  coincide  with  face  CD,  etc. 

537.  THEOREM.  Two  vertical  dihedral  angles  are  equal.  (See  535.) 

538.  THEOREM.     The  plane  angle  of  a  right  dihedral  angle  is  a 
right  angle;  and  if  the  plane  angle  of  a  dihedral  angle  is  a  right  angle, 
the  dihedral  angle  is  right.     (See  531.) 

539.  THEOREM.     Two  dihedral  angles  have  the  same  ratio  as  their 
plane  angles. 

Given  :     Dihedral  A  A-BC-D  and  Ar-Brc'-Df,  having  plane 


A  ACE  and  A'C'E',  respectively. 

Dih.  Z  A-BC-D  :  dih.  Z  A'-BfCf-Dr=  Z  ACE 

Proof:  I.  If  the  plane  angles 
are  commensurable, 

There  exists  a  common  unit  of 
measure  of  the  plane  A  ACE  and 
A'C'E*  (?)  (238).  Suppose  this 
unit  when  applied  to  these  angles 
is  contained  3  times  in  Z  ACE 
and  4  times  in  Z  A'C'E'. 


To  Prove: 

^  A'C'E'. 


cf 


NN: 

\: 


JL 


Pass  planes  through  the  edges 
and   the    several    lines    of    division    of    the    angles.      Dih. 
Z  A-BC-D  is  divided  into  3  parts ;  dih.  Z  Af-BrCf-Dr  is  di- 
vided into  4  parts ;  all  of  these  seven  parts  are  equal  (?)  (535). 

dih.  Z  A-BC-D     _  3  ^  ,Ax   QN 


/.dih.  Z  A-BC-D:  dih. 


276 


SOLID   GEOMETRY 


II.  If  the  plane  angles  are 
incommensurable. 

There  does  not  exist  a  com- 
mon unit  (?).  Suppose  Z  ACE 
to  be  divided  into  equal  parts 
(any  number  of  them). 

Apply  one  of  these  as  a 
unit  of  measure  to  Z  A'c'E1 . 


B 


There  will  be  a  remainder, 

,  left  over  (because  the  A  are  incommensurable). 

Pass  a  plane  C'F,  determined  by  B!C'  and  CrX. 
dih.  Z  A-BC-D 


Now 


(Commensurable  A). 


dih.  z  A'-B'C'-Y    Z.A'C'X 
Indefinitely  increase  the  number  of  subdivisions  of  Z.ACE. 

Then  each  part,  that  is,  our  unit  or  divisor,  will  be  indefi- 
nitely decreased.  Hence  XC'  Er,  the  remainder,  will  be  in- 
definitely decreased. 

That  is,  Z  XC'E'  will  approach  zero  as  a  limit  ;  and  dih. 
Z.X-BrCr-D'  will  approach  zero  as  a  limit. 


'x  will  approach  A'C'E'  as  a  limit  (240);  and  dih. 
Z  A'-B'c'-Y  will  approach  dih.  Z  A'-B'C'-D'  as  a  limit  (240). 

dih. 


dih.  /.A'-B'C'-Y 

v     .,        -,     Z.ACE       .-,,  ,      Z.  ACE 

limit  and  -      r— r—  will  approach  r~r~7  as  a  limi^ 


/.A'C'X 

dih.  Z  A-BC-D 


Z  ACE 


dih.  Z  A'-B'C'-D'     /.A'C'E' 


(242). 


Q.E.D. 


BOOK  VI 


277 


540.  THEOREM.   If  a  straight  line  is  perpendicular  to  a  plane,  any 
plane  containing  this  line  is  perpendicular  to  the  given  plane. 

r-k 

Given  :  Line  AB  _L  to  plane  MN-, 
plane  PQ  containing  AB  and  inter- 
secting plane  MN  in  RS. 

To  Prove :    PQ  is  _L  to  MN. 

Proof  :    In  MN  draw  BC  _L  to  US. 

AB  is  ±  to  BS  and  to  BC  (?)  (489). 

.-.  </ABC  is  the  plane  Z  of  dih.  £P-8B-N  (?)  (530). 

But  Z  ABC  is  a  rt.  Z  (?). 

.-.  PQ  is  _L  to  MN  (?)  (538).  Q.E.D. 

541.  THEOREM.     If  a  plane  is  perpendicular  to  the  edge  of  a  di- 
hedral angle,  it  is  perpendicular  to  each  face.     (See  540.) 

542.  If  one  plane  is  perpendicular  to  another,  any  line  in  either  plane, 
perpendicular  to  their  intersection,  is  perpendicular  to  the  other  plane. 

Given  :    Plane  PQ  _L  to  plane  MN ;  AB  in  plane  PQ  _L  to  the 
intersection,  RS. 

To  Prove :  AB  _L  to  plane  MN. 

Proof :  In  plane  MN  draw  BC  J_  to  RS. 

Now,  /_ABC  is  the  plane  angle  of  the  dih.  Z  P-SR-N  (?). 

/.  Z  ABC  is  a  rt.  Z  (?)  (538).     .'.AB  is±to  BC  (?)  (17). 

But  AB  is  _L  to  RS  (Hyp.). 

/.  AB  is  _L  to  plane  MN  (?)  (501).  Q.E.D. 


Ex.  1.  In  the  figure  of  540  prove  BC  perpendicular  to  plane  PQ. 
Also  prove  RS  perpendicular  to  the  plane  ABC. 

Ex.  2.  In  the  figure  of  540  prove  the  plane  ABC  perpendicular  to 
pianes  MN  and  PQ. 

Ex.  3.  Under  what  condition  will  a  line  in  one  face  of  a  dihedral 
angle  meet  a  line  in  the  other  face  ? 


278 


SOLID  GEOMETRY 


543.  THEOREM.  If  one  plane  is  perpendicular  to  another,  a  line 
drawn  from  any  point  in  their  intersection  and  perpendicular  to  one 
plane,  lies  in  the  other. 

Given  :  Plane  PQ  ±  to  plane  MN, 
intersecting  in  BS ;  AB  _L  to  plane 
MN  from  A,  in  US. 

To  Prove :    AB  is  in  plane  PQ. 
Proof  :    At  A  erect  in  plane  PQ, 
AX  ±  to  R8. 

Then  AX  is  J_  to  plane  MN  (?)  (542). 

But  AB  is  _L  to  plane  MN  at  A  (Hyp.). 

.-.  AB  and  AX  coincide  (?)  (506). 

That  is,  AB  is  in  plane  PQ.  Q.E.D. 

544.  THEOREM.     If  one   plane  is  perpendicular  to  another,  a  line 
drawn  from  any  point  in  one  plane,  and  perpendicular  to  the  other,  lies 
in  the  first  plane. 

Given  :    Plane  PQ  J_  to  plane  MN ;  AB  _L  to  plane  MN  from 
A,  any  point  in  plane  PQ. 
To  Prove:  (?). 

Proof :  From  A  draw  in  plane  PQ,  AX  _L  to  ES. 
Then  AX  is  _L  to  plane  MN  (?)  (542). 
/.  AB  and  AX  coincide  (?).  Etc. 

545.  THEOREM.     If  two  planes  are  perpendicular  to  a  third  plane, 
their  intersection   also    is   perpen- 
dicular to  that  plane. 

Given:    Planes  LM  and  NP, 
each  _L  to  plane  R S. 

To  Prove  :    The   intersection 
AB  is  -L  to  plane  BS. 

Proof:  If,    at   A,   a   line   be  ™S 

erected  JL  to  plane  BS,  it  will  lie  in  plane  LM  (?)  (543). 
This  _L  will  lie  also  in  plane  NP  (?). 
/.  this  J_  is  the  intersection  AB  (?)  (482). 
That  is,  AB  is  _L  to  plane  BS.  Q.E.D. 


BOOK  VI  279 

546.  THEOREM.     If  a  plane  is  perpendicular  to  each  of  two  inter- 
secting planes,  it  is  perpendicular  to  their   intersection.    (The  same 
truth  as  545.) 

547.  THEOREM.     If  each  of  three  planes  is  perpendicular  to  the 
other  two,  each  of  the  three   intersections  is  perpendicular  to  the 
remaining  plane,  and  perpendicular  to  the  other  two  intersections. 

That  is,  if  each  of  the  three  planes  RS,  LM,  NP  is  _L  to  the 
others,  then, 

BA  is  _L  to  plane  ES,  to  LA,  and  to  NA; 


LA  is  JL  to  plane  NP,  to  AB,  and  to  NA; 


(545  and  489). 


NA  is  J_  to  plane  LM,  to  AB,  and  to  LA. 

This  truth  is  illustrated  at  the  corner  of  an  ordinary  box  or  room. 

548.  THEOREM.     Through  a  given  line  not  perpendicular  to  a  plane, 
one  plane  can  be  passed  perpendicular  to  that  plane,  and  only  one. 

Given:  AB  not  _1_  to  plane  MN. 

To  Prove :  Through  AB  one  plane 
can  be  passed  J_  to  MN,  and  only  one. 

Proof :  I.  From  P,  any  point  in 
AB,  draw  PX  J.  to  MN.  Through  A  B 
and  PX  pass  plane  AC.  Plane  AC  is  _L  to  plane  MN  (?)  (540). 

II.  Suppose  another  plane  containing  AB  is  _L  to  plane  MN. 
Then  the  intersection  AB,  of  these  two  planes,  which  are  _L  to 
plane  MN,  will  be  _L  to  plane  MN  (?)  (545). 

But  AB  is  not  JL  to  plane  MN  (?)  (Hyp.). 

/.  there  is  only  one  plane  containing  AB  and  _L  to  plane  MN. 

Q.E.D. 

549.  COR.     The  plane  containing  a  straight  line  and  its  projection 
upon  a  plane  is  perpendicular  to  the  given  plane. 


280 


SOLID   GEOMETRY 


550.   THEOREM.     One  line  can  be  drawn  perpendicular  to  both  or 
two  lines  in  space,  not  in  the  same  plane,  and  only  one. 

Given:  Lines  AB  and  CD  not  in  the  same  plane. 

To  Prove:   One  line  can  be  drawn  JL  to  AB  and  CD,  and 
only  one. 

Proof:  I.  At  P,  any  point 
in  CD,  draw  EF  II  to  AB.  Pass 
plane  JOT,  containing  CD  and 
EF.  Pass  plane  AH  through 
AB  and  _L  to  JOT,  intersecting 
MN  in  GH,  and  CD  at  L.  In 
plane  AH  draw  EL  _L  to  GH. 


RL  is  _L  to  plane  MN  (?)(542). 
(?)  (95). 


Plane  MN  is  II  to  AB  (498). 
GH  is  II  to  AB  (?)  (499). 
EL  is  JL  to  CH  (Const.).     .* 

.-.  EL  is  -L  to  CD  (?).     Also,  EL  is  _L  to 
That  is,  EL  is  J_  to  both  the  given  lines. 

II.  If  another  line  can  be  drawn  _L  to  AB  and  CD,  suppose 
SP  is  this  J_.  In  plane  AH  draw  sx  J_  to  GH.  Then  8X  is 
JL  to  plane  MN  (?)  (542). 

But  if  SP  is  _L  to  AB,  it  is  _L  to  EF  (?)  (95). 

.-.  SP  is_L  to  plane  MN  (?)  (501). 

Thus  there  are  two  J§  from  S  to  plane  MN  (-SJT  and  SP). 

But  this  is  impossible  (?)  (507). 

/.  there  can  be  no  second  JL  to  these  two  given  lines. 

Q.E.D. 


Ex.  1.    In  the  figure  of  550  prove  that  a  plane  perpendicular  to  PL 
at  its  midpoint  will  be  parallel  to  AB  and  CD. 
Ex.  2.   Prove,  also,  that  this  plane  will  bisect  SP. 
Ex.  3.     Prove  that  if  CD  is  not  perpendicular  to  EF,  no  plane  can  be 
through  AB,  perpendicular  to  CD* 


BOOK   VI 


281 


551.  THEOREM.     Every  point  in  a  plane  bisecting  a  dihedral  angle 
is  equally  distant  from  the  faces  of  the  angle. 

Given:  Plane  AB,  bisecting 
the  dih.  Z  C-BD-E ;  any  point 
P  in  plane  AB ;  PF  _L  to  face 
CB  ;  PH  JL  to  face  DE. 

To  Prove:  PF=PH. 

Proof:  Pass  plane  MN,  con- 
taining PF  and  PH,  intersect- 
ing CB  in  FG,  AB  in  PG,  DE  in 

xzGr«    /j  Jj  £tu    Cr« 

Now  plane  MN  is  _L  to  planes  CB  and  D#  (?)  (540). 
.*.  plane  MN  is  -L  to  JBD  (?)  (546). 
.-.  BG  is  J_  to  FG,  PG,  and  #£  (?)  (489). 

Hence,  Z  P£F  is  the  plane  Z  of  dih.  Z  A-BD-C  and  Z 
is  the  plane  Z  of  dih.  Z  A-BD-E  (?)  (530). 

These  dih.  ^  are  =  (Hyp.).    .'.Z  p#p  =  Zp#ff  (?)  (536). 
Z  PPG  and  PffG  are  rt.  A  (?)  (489). 

In  the  right  A  PFG  and  PHG,  PG  =  PG  (?), 
and  Z  PGF=  Z  PGZT  (?). 

/.  these  A  are  =  (?).    Consequently,  PF—  PH  (?).   Q.E.D. 

552.  THEOREM.      Any  point   in   a   dihedral   angle   and    equally 
distant  from  the  faces  of  the  angle  is  in  the  plane  bisecting  the 
angle. 

Given:  PF=PH. 

To  Prove:    Plane  AB,  determined  by  P  and  DB,  is    the 
bisector  of  dih.  Z  C-BD-E. 

Proof :  Similar  to  the  proof  of  551. 

\ 

553.  THEOREM.     The  plane  bisecting  a  dihedral  angle  is  the  locus 
of  points  in  space  equally  distant  from  the  faces  of    the  dihedral 
angle.     (See  551,  552.) 


282  SOLID  GEOMETRY 

ORIGINAL   EXERCISES 

1.  Are    two    planes  perpendicular    to    the  same  plane  necessarily 
parallel? 

2.  A  straight  line  and  a -plane  perpendicular  to  the  same  plane  are 
parallel. 

3.  A  plane  perpendicular  to  a  line  in  another  plane,  is  perpendicular 
to  that  plane. 

4.  If  three  planes,  all  perpendicular  to  a  fourth,  intersect  in  three 
lines,  these  lines  are  parallel,  in  pairs. 

5.  If  the  projection  of  any  line  (straight  or  curved)  upon  a  plane  is 
a  straight  line,  the  line  is  entirely  in  one  plane. 

6.  The  angle  between  the  normals  drawn  to  the  faces  of  a  dihedral 
angle  from  a  point  within   the  angle  is  the  supplement   of   the  plane 
angle  of  the  dihedral  angle. 

7.  If  a  line  is  parallel  to  a  plane,  any  plane  perpendicular  to  the  line 
is  perpendicular  also  to  the  plane. 

[Construct  the  projection  of  the  given  line  upon  the  given  plane.] 

8.  What  is  the   locus  of  points  in  space  equally  distant  from  two 
intersecting  planes? 

9.  If  from  any  point  in  a  face  of  a  dihedral 
angle,  a  normal  is  drawn  to  each  face,  the  plane  of 
these  normals  is  perpendicular  to  the  edge  of  the        M 
dihedral  angle. 

10.  If  from  any  point  in  a  face  of  a  dihedral 
angle,  a  normal    is  drawn  to  each   face,  the  angle 
they  form    is   equal   to    the    plane   angle    of    the 
dihedral  angle. 

11.  If  a  line  is  perpendicular  to  a  plane,  any  plane  parallel  to  the 
line  is  also  perpendicular  to  the  plane. 

12.  If   PA   is   a  normal  to  plane  MN,  PB  a       M. 
normal  to  plane   ST,  and  BC  a  normal  to  plane          / 
MN,  AC  is  perpendicular  to  RS,  the  intersection        L 
of  planes  MN  and  ST. 

13.  The  plane  perpendicular  to   the  line  that  is  perpendicular  to 
two  lines  in  space,  at  its  middle  point,  bisects  every  straight  line  having 
its  extremities  in  these  lines. 

14.  The  common  perpendicular  to  two  lines  in  space  is  the  shortest 
line  that  can  be  drawn  between  them. 


BOOK   VI 


283 


15.  The  plane  perpendicular  to  the  plane  of  an 
angle  and  containing  the  bisector  of  the  angle  is 
the  locus  of  points  equally  distant  from  the  sides 
of  the  angle. 

Proof:  The  J§  from  any  point  in  plane  NR  to 
AB  and  EC  will  have  equal  projections.    (Explain  by  use  of  79.) 
.•.  these  perpendiculars  are  equal  (?). 

16.  What  is  the  locus 'of  points  in  space  equally  distant  from  two 
intersecting  lines  ? 

17.  If  A  P  is  a  normal  to  plane  MN   and    if 
angle  PBC,  in  plane  MN,  is  a  right  angle,  angle 
ABC  also  is  a  right  angle. 

[Prove  BC  is  _L  to  plane  APE]. 
18.    If  AP  is  a  normal  to  plane  MN  and  Z.  PBD, 
in  plane  MN,  is  obtuse,  /.+ 4  BD  also  is  obtuse. 


M, 


N 


Proof:  Take  BC  =  BD-,  prove  PD>  PC.    Then  prove  AD>AC,  etc. 

19.  PA  is  perpendicular  to  plane  72S;  AB  and 
PC  are  perpendicular    to  plane    MR.      Prove    BC 
perpendicular  to  RT. 

20.  If  two  parallel  planes  are  cut  by  a  third  plane, 
the  alternate-interior  dihedral  angles  are  equal;  the 
corresponding  dihedral  angles  are  equal ;  the  alternate- 
exterior  dihedral  angles  are  equal ;  the  adjoining  in- 
terior dihedral  angles  are  supplementary. 


21.    State  and  prove  the  converse  theorems  of  those  in  No.  20. 


22.  To  construct  a  plane  perpendicular  to  a  given  plane  and  contain- 
ing a  given  line  in  that  plane. 

23.  To  construct  a  plane  perpendicular  to  a  given  plane  and  contain- 
ing a  given  line  without  that  plane. 

24.  To  construct  through  a  given  point  a  line  which  will  intersect 
any  two  given  lines  in  space. 

Construction :  Pass  a  plane  through  the  point  and  one  of  the  lines. 
This  plane  intersects  the  other  line  at  a  point.  This  point  and  the 
given  point  determine  the  required  line.  Explain.  Discuss. 

25.  To  bisect  a  given  dihedral  angle. 

Construction :  Pass  a  plane  JL  to  the  edge.  '  Bisect  the  plane  Z 
formed,  pass  the  plane  determined  by  this  bisector  and  the  given  edge. 


284 


SOLID   GEOMETRY 


26.  To  construct  a  line  whose  points  shall  be  equally  distant  from 
the  ends  of  a  given  line  and  also  equally  distant  from  the  faces  of  a 
dihedral  angle. 

27.  To  find  the  locus  of  points  equally  distant  from  two  points  and 
also  equally  distant  from  two  intersecting  planes.     Discuss. 

28.  To   find  a  point  equally  distant   from    three  given  points  and 
equally  distant  from  two  intersecting  planes. 

Is  this  problem  ever  impossible?     Will  there  ever  be  two  points? 
When  will  there  be  only  one  ? 

29.  To  find  a  point  equally  distant  from  three  given   points  and 
equally  distant  from  two  intersecting  lines.     Discuss  fully. 


POLYHEDRAL  ANGLES 

554.  If  three  or  more  planes  meet  at  a  point,  they  form  a 
polyhedral  angle.     The  opening  partially  surrounded  by  the 
planes  is  the  polyhedral  angle. 

The  point  common  to  all  the  planes  is  the  vertex. 

The  planes  are  the  faces. 

The  intersections  of  adjacent  faces  are  the  edges. 

The  angles  formed  at  the  vertex,  by  adjacent  edges,  are 
the  face  angles. 

Thus,  V—  ABODE  is  a  polyhedral 
angle  ;  V  is  the  vertex ;  A  V,  BV,  etc., 
are  edges;  planes  AVB,  BVC,  etc., 
are  faces;  A  A  VB,  BVC,  etc.,  are 
face  angles. 

555.  A   plane   section    of    a 

polyhedral    angle    is    the   plane  A 
figure  bounded  by  the  intersec- 
tions of  all  the  faces  by  a  plane. 

Polygon  LMNOP  is  a  plane  section  of  polyhedral  angle  V— ABODE. 


A  convex  polyhedral  angle  is  one  whose  plane  sections  are 
all  convex. 


EQUAL  POLYHEDRAL 
ANGLES 


VERTICAL 

POLYHEDRAL 

ANGLES 


VERTICAL 

DIHEDRAL 

ANGLES 


SYMMETRICAL 

POLYHEDRAL 

ANGLES 


556.  Two  polyhedral  angles  are  equal  if  they  can  be 
made  to  coincide  in  all  particulars.  That  is,  if  two  polyhe- 
dral angles  are  equal,  their  homologous  dihedral  angles  are 
equal;  their  homologous  face  angles  are  equal,  and  they  are 
arranged  in  the  same  order.  The  length  of  the  edges  or 
the  extent  of  the  faces  does  not  affect  the  size  of  the  poly- 
hedral angle. 

Two  polyhedral  angles  are  vertical  if  the  edges  of  one  are 
the  prolongations  of  the  edges  of  the  other. 

Two  polyhedral  angles  are  symmetrical  if  all  the  parts  of 
one  are  equal  to  the  corresponding  parts  of  the  other,  but 
arranged  in  opposite  order. 

NOTE.  It  is  apparent  from  the  definitions  that  equal  polyhedral 
angles  are  mutually  equiangular,  as  to  the  face  angles  and  as  to  the  dihe- 
dral angles. 

Vertical  polyhedral  angles  are  mutually  equiangular,  as  to  their  face 
angles  and  as  to  their  dihedral  angles,  but  the  order  is  reversed. 

Symmetrical  polyhedral  angles  are  also  mutually  equiangular  as  to 
their  face  angles  and  as  to  their  dihedral  angles,  but  the  order  is 
reversed. 

Thus,  if  one  follows  around  the  polygon  A'D'  in  alphabetical  order, 
he  is  moving  as  the  hands  of  a  clock  —  if  the  eye  is  at  the  vertex  0'; 
but  if  he  follows  around  AD  alphabetically,  he  is  moving  in  a  direction 
opposite  to  the  motion  of  the  hands  of  a  clock  —  if  the  eye  is  at  the 
Yertex  0. 

Hence,  it  is  apparent  that,  in  general,  symmetrical  polyhedral  angles 
cannot  be  made  to  coincide. 


286  SOLID   GEOMETRY 

557.  A  trihedral  angle  is  a  polyhedral  angle  having  three 
and  only  three  faces. 

A  trihedral  angle  is  rectangular  if  it  contains  a  right  dihe- 
dral angle;  birectangular,  if  it  contains  two  right  dihedral 
angles;  trirectangular,  if  it  contains  three  right  dihedral 
angles. 

A  trihedral  angle  is  isosceles  if  two  of  its  face  angles  are 
equal. 

558.  THEOREM.    Two  vertical  polyhedral  angles  are  symmetrical. 

Proof :  Their  homologous  face  angles  are  equal  and 
arranged  in  reverse  order,  and  their  homologous  dihedral 
angles  are  equal  and  arranged  in  reverse  order. 

/.they  are  symmetrical  (Def.  556). 

559.  THEOREM.     If  two  polyhedral  angles  are  symmetrical,  the 
vertical  polyhedral  angle  of  the  one  is  equal  to  the  other. 

Because  the  corresponding  parts  are  equal  and  they  are 
arranged  in  the  same  order. 

560.  THEOREM.   Provided  two  trihedral  angles  have  their  parts 
arranged  in  the  same  order,  they  are  equal : 

I.  If  two  face  angles  and  the  included  dihedral  angle  of  one  are 
equal  respectively  to  two  face  angles  and  the  included  dihedral  angle 
of  the  other. 

II.  If  a  face  angle  and  the  two  dihedral  angles  adjoining  it  of  the  one 
are  equal  respectively  to  a  face  angle  and  the  two  dihedral  angles 
adjoining  it,  of  the  other. 

Proof :  By  method  of  superposition,  as  in  plane  A. 


BOOK   VI 


287 


561.  THEOREM.  Provided  two  trihedral  angles  have  their  parts 
arranged  in  the  same  order,  they  are  equal,  if  the  three  face  angles  of 
one  are  equal  respectively  to  the  three  face  angles  of  the  other. 

Given:  Trih.  A  O  and  o'; 

.  /  Atn'R'  •  O  O' 


C'O'A'. 


To  Prove:  Trih.  Z  O  = 
trih.  Z  o',  that  is,  dih.  Z  OA 
=  dih.  Z  O'A',  etc. 

Proof :   Take  OA  =  OB  =  oc  =  O'A'  =  O'B'  =  o'c'. 

Draw  AB,  BC,  AC,  A'B',  B'C',  A'c'. 

Take,  on  edges  AO  and  A'O',  AP  =  A'P'  and  in  face  AOB 
draw  PD  _L  to  AO. 

Z  OAB  is  acute  (A  A  OB  is  isosceles).     .-.  PD  will  meet  AB. 

In  face  AOC  draw  PE  _L  to  AO,  meeting  .4 Cat  E.     Draw  DE. 
Similarly  draw  P'D',  P' E',  D'E' . 

.  Now  A  EPD  and  E'P'D'  are  the  plane  A  of  the  dihedral  A 
AO  and.  A1 o'  (?)  (530).  To  prove  that  these  A  are  equal 
requires  the  proof  that  eight  pairs  of  A  are  equal. 

AB=A'B',  BC=B'C', 


(1)  A  OAB  =  A  O'A'B'.   (Explain.) 

(2)  A  OBC=A  O'B'C'.   (Explain.) 

(3)  A  OAC  =  A  o'A'c'.   (Explain.) 


AC  =  A'C',    Z    OAB  = 

Z  O'A'B',  Z  OAC  = 
Z  O'A'C'  (?)  (27). 

•.     AD  =  A'D',     AE  = 
!'#';  PD  =  P'l/,etc.(?). 
/r'  A'  R'  fv\ 

Z.C/  A  K    {.  J. 

=  E'D'  (?). 

=  Z^'P'D'(?). 
Hence,  dih.  Z  ^4O  =  dih.  Z  A'O'  (?)  (535). 
Similarly  one  may  prove  the  other  pairs  of  homologous 
dihedral  angles  equal. 

.'.  trihedral  Z  O  =  trihedral  Z  o'  (?)  (556).  Q.E.D. 


(4)  A^PD  =  A^L'P'D'.  (Explain.) 

(5)  A!P^=A^/P'^/.  (Explain.) 

(6)  A  ABC  =  A  A' B'C'.  (Explain.) 

(7)  A  AED  =  A  A'E'D'.  (Explain.) 

(8)  A  PED=  A  P'E'D'.  (Explain.) 


288 


SOLID   GEOMETRY 


562.  THEOREM.    Provided  two  trihedral  angles  have  their  parts 
arranged  in  reverse  order,  they  are  symmetrical : 

I.  If  two  face  angles  and  the  included  dihedral  angle  of  one  are 
equal  respectively  to  two  face  angles  and  the  included  dihedral  angle  of 
the  other. 

II .  If  a  face  angle  and  the  two  dihedral  angles  adjoining  it  of  the 
one  are  equal  respectively  to  a  face  angle  and  the  two  dihedral  angles 
adjoining  it  of  the  other. 

III.  If  the  three  face  angles  of  one  are  equal  respectively  to  the  three 
face  angles  of  the  other. 

Proof :  In  each  case  construct  a  third  trihedral  Z  symmet- 
rical to  the  first.  This  third  figure  will  have  its  parts  =  to  the 
parts  of  the  second,  and  arranged  in  the  same  order  (Def .  556). 

.-.the  third  =  the  second  (?)  (560  and  561). 

/.  the  first  is  symmetrical  to  the  second  (Ax.  6).       Q.E.D. 

563.  THEOREM.   The  sum  of  any  two  face  angles  of  a  trihedral  angle 
is  greater  than  the  third  face  angle. 

Given:   Trih.  Z.O-RST\\\  which 
face  angle  ROT  is  the   greatest. 
To  Prove: 

Z    EO-S  +  Z    SOT   >  Z  ROT. 

Proof :    Construct,  in  face  KOT, 
Z  ROD  =  Z  ROS. 

Take  OD=OB\  draw  ADC,  meet- 
ing OT  at  C.     Draw  AB  and  BC. 

A  AOD  =  A  AOB.     (Explain.) 

.:AB  =  AD(?). 

Now  AB  +  BC  >  AD  +  DC  (?)  (Ax.  12). 

But,  AB  =  AD  (?). 

Subtracting,  BC  >  DC  (Ax.  7). 

Now,  OB  =  OD  (?),  OC  =  OC  (?)  and  BC  >  DC  (Just  proved). 
.-.  Z  BOG      >  Z  DOC  (?)  (87). 

But  Z  AOB =  Z  AOD  (?). 

Adding,  Z  AOB  +  BOC      >  Z  AOC  (Ax.  7). 

That  is,  Z  ROS  +Z  SOT  >  Z  ROT  (?)  (Ax.  6).        Q.E.D. 


BOOK  VI  289 

564.  THEOREM.  The  sum  of  the  face  angles  of  any  polyhedral  angle 
is  less  than  four  right  angles,  or  360°. 

Given:  Polyhedral  Z  O,  having  n  faces. 

To  Prove :  The  sum  of  the  face  A  at  O 
<  4rt.  4  or  360°. 

Proof :  Pass  a  plane  AD,  intersecting  all 
the  faces,  and  the  edges  at  A,  B,  C,  etc. 

In  this  section  take  any  point  X  and 
join  X  to  all  the  vertices  of  the  polygon.  |B 

(1)  There  are  n  face  A  having  their  vertices  at  O  (Hyp.). 

(2)  There  are  n  base  A  having  their  vertices  at  X  (Const.). 

(3)  The  sum  of  the  A  of  the  face  A  =  2  n  rt.  A  (110). 

(4)  The  sum  of  the  A  of  the  base  A  =  2  n  rt.  A  .(110). 

(5)  .-.  the  sum  of  the  A  of  the  face  A  =  the  sum  of  the  A 
of  the  base  A  (Ax.  1.). 

Now,  Z  OAE  +  Z  OAB  >  Z.EAB  (?)  (563). 
And   Z  OBA  +  Z  OBC  >  /-ABC  (?),  etc.,  etc. 

Adding,  the  sum  of  the  base  A  of  the  face  A  >  the  sum  of 
the  base  A  of  the  base  A  (Ax.  8). 

Subtracting  this  inequality  from  equation  (5)  above,  the 
sum  of  the  face  A  at  O  <  the  sum  of  the  A  at  X  (Ax.  9). 

But  the  sum  of  all  the  A  at  X=  4  rt.  A  (?)  (47). 

.-.the  sum  of  the  face  A  at  O  <  4  rt.  A,  or  360°  (Ax.  6). 

Q.E.D. 

Ex.  1.   Prove  theorem  of  563  for  the  case  of  an  isosceles  trihedral  angle. 

Ex.  2.  In  the  figure  of  564,  as  the  vertex  O  approaches  the  base,  does 
the  sum  of  the  face  angles  at  O  increase  or  decrease?  What  limit  does 
this  sum  apDroach?  Does  the  sum  ever  become  equal  to  this  limit? 


\  \ 


290  SOLID  GEOMETRY 

ORIGINAL   EXERCISES 

1.  The  three  planes  bisecting  the  three  dihedral  angles  of  a  trihedral 
angle  intersect  in  a  straight  line. 

2.  The  three  planes  containing  the  three  bisectors  of  the  three  face 
angles  of  a  trihedral  angle  and  perpendicular  to  those  faces  intersect  in 
a  straight  line. 

3.  If  two  face  angles  of  a  trihedral  angle   are 
equal,  the  dihedral  angles  opposite  them  are  equal. 

Given:  /.RVS=£SVT. 
To  Prove:    Dih.  Z  R  V=  dih.  Z  TV. 
Proof:    Pass   plane  SVX   bisecting   dih.  Z   SV. 
Prove  trih.  A  V-RSX  and  V-TSX  are  sym.  by  562, 1. 

4.  An  isosceles  trihedral  angle  and  its  symmet- 
rical trihedral  angle  are  equal. 

5.  Find  the  locus  of  points  equally  distant  from  the  three  faces  of  a 
trihedral  angle. 

6.  Find  the  locus  of  points  equally  distant  from  the  three  edges  of 
a  trihedral  angle. 

7.  If  the  three  face  angles  of  a  trihedral  angle  are  equal,  the  three 
dihedral  angles  also  are  equal. 

8.  If  the  three  face  angles  of  a  trihedral  angle  are  right  angles,  the 
three  dihedral  angles  also  are  right  angles. 

9.  In  any  trihedral  angle  the  greatest  dihedral  angle  has  the  greatest 
face  angle  opposite  it. 

10.  If  the  edges  of  one  trihedral  angle  are  perpendicular  to  the  faces 
of  a  second  trihedral  angle,  then  the  edges  of  the  second  are  perpen- 
dicular to  the  faces  of  the  first. 

11.  To  construct  the  plane  angles  of  the  three  dihedral  angles  of  a 
trihedral  angle,  if  the  three  face  angles  are  known. 

[To  be  accomplished  by  constructions  in  a  plane.] 

12.  To  construct,  through  a  given  point,  a  plane  which  shall  make, 
with  the  faces  of  a  polyhedral  angle  having  four  faces,  a  section  that 
is  a  parallelogram. 

Construction:  Extend  one  pair  of  opp.  faces  to  obtain  their  line  of 
intersection.  Similarly  extend  the  other  pair.  Any  plane  section  ||  to 
these  lines  will  be  a  £7.  (Explain.) 


BOOK  VII 

POLYHEDRONS 

565.   A  polyhedron  is  a  solid  bounded  by  planes. 

The  edges  of  a  polyhedron  are  the  intersections  of  the 
bounding  planes. 

The  faces  are  the  portions  of  the  bounding  planes  included 
by  the  edges. 

The  vertices  are  the  intersections  of  the  edges. 

The  diagonal  of  a  polyhedron  is  a  straight  line  joining  two 
vertices  not  in  the  same  face. 


POLY-     TETKA-    HEXAHEDRON    OCTA-       DODECA-      ICOSA- 
HEDRON   HEBRON       CUBE        HEDRON     HEDRON     HEBRON 

566.  A  tetrahedron  is  a  polyhedron  having  four  faces. 
A  hexahedron  is  a  polyhedron  having  six  faces. 

An  octahedron  is  a  polyhedron  having  eight  faces. 
A  dodecahedron  is  a  polyhedron  having  twelve-  faces. 
An  icosahedron  is  a  polyhedron  having  twenty  faces. 

567.  A  polyhedron  is  convex  if  the  section  made  by  every 
plane  is  a  convex  polygon. 

Only  convex  polyhedrons  are  considered  in  this  book. 

291 


292  SOLID   GEOMETRY 

PRISMS 

568.  A  prism  is  a  polyhedron  two  of  whose  opposite  faces 
are  equal  polygons  in  parallel  planes,  and  whose  other  faces 
are  all  parallelograms. 

The  bases  of  a  prism  are  the  equal,  parallel  polygons. 

The  lateral  faces  of  a  prism  are  the  parallelograms. 

The  lateral  edges  of  a  prism  are  the  intersections  of  the 
lateral  faces. 

The  lateral  area  of  a  prism  is  the  sum  of  the  areas  of  the 
lateral  faces. 

The  total  area  of  a  prism  is  the  sum  of  the  lateral  area  and 
the  areas  of  the  bases. 

The  altitude  of  a  prism  is  the  perpendicular  distance  be- 
tween the  planes  of  the  bases. 

A  triangular  prism  is  a  prism  whose  bases  are  triangles. 


PRISM  TRIANGULAR      REGULAR      OBLIQUE  PRISMS         RIGHT  SECTION 

PRISM  PRISM  TRUNCATED  PRISMS 

569.  A  right  prism  is  a  prism  whose  lateral  edges  are  per- 
pendicular to  the  planes  of  the  bases. 

A  regular  prism  is  a  right  prism  whose  bases  are  regular 
polygons. 

An  oblique  prism  is  a  prism  whose  lateral  edges  are  not  per- 
pendicular to  the  planes  of  the  bases. 

A  truncated  prism  is  the  portion  of  a  prism  included  between 
the  base  and  a  plane  not  parallel  to  the  base. 


BOOK  VII  293 

A  right  section  of  a  prism  is  the  section  made  by  a  plane 
perpendicular  to  the  lateral  edges  of  the  prism. 

570.  A  parallelepiped  is  a  prism  whose  bases  are  parallelo- 
grams. 

A  right  parallelepiped  is  a  parallelepiped  whose  lateral  edges 
are  perpendicular  to  the  planes  of  the  bases. 

A  rectangular  parallelepiped  is  a  right  parallelepiped  whose 
bases  are  rectangles. 


PARALLELEPIPED  RIGHT          RECTANGULAR      CUBE 

PARALLELEPIPED   PARALLELEPIPED 

An  oblique  parallelepiped  is  a  parallelepiped  whose  lateral 
edges  are  not  perpendicular  to  the  planes  of  the  bases. 

A  cube  is  a  rectangular  parallelepiped  whose  six  faces  are 
squares. 

571.  The  unit  of  volume  is  a  cube  whose  edges  are  each  a 
unit  of  length. 

The  volume  of  a  solid  is  the  number  of  units  of  volume  it 
contains.  The  volume  of  a  solid  is  the  ratio  of  that  solid  to 
the  unit  of  volume. 

The  three  edges  of  a  rectangular  parallelepiped  meeting  at 
any  vertex  are  the  dimensions  of  the  parallelepiped. 

Equivalent  solids  are  solids  that  have  equal  volumes. 

Equal  solids  are  solids  that  can  be  made  to  coincide. 


Ex.   What  is  the  base  of  a  rectangular  parallelepiped?    Of  a  right 
parallelepiped  ?    Of  an  oblique  parallelepiped  ? 


294  SOLID  GEOMETRY 

NOTE.  The  space  that  is  bounded  by  the  surfaces  of  a  solid,  independ- 
ent of  the  solid,  is  called  a  geometrical  solid. 

That  is,  if  a  material  or  physical  body  occupy  a  certain  position  and 
then  be  removed  elsewhere,  there  is  a  definite  portion  of  space  that  is 
the  exact  shape  and  size  as  the  solid,  and  can  be  conceived  as  bounded 
by  exactly  the  same  surfaces  as  bounded  the  solid  when  in  that  original 
position.  In  order  that  we  may  pass  planes  and  draw  lines  through 
solids,  and  superpose  one  solid  upon  another,  it  is  convenient  in  studying 
the  properties  of  solids  to  consider  them  usually  as  geometric  solids,  — 
the  material  body  being  removed  for  the  time. 

The  three  kinds  of  parallelepipeds  can  be  illustrated  by  removing 
the  cover  and  bottom  of  an  ordinary  cardboard  box,  and  distorting  the 
shape  of  the  frame  that  remains. 


PRELIMINARY  THEOREMS 

572.  THEOREM.      The  lateral  edges  of  a  prism  are  all  equal  (?). 

573.  THEOREM.      Any  two  lateral  edges  of  a  prism  are  parallel. 
(See  511.) 

574.  THEOREM.     Any  lateral   edge  of  a  right   prism  equals  the 
altitude.     (See  524.) 

575.  THEOREM.     The  lateral  faces  of  a  right  prism  are  perpen- 
dicular to  the  bases.     (See  540.) 

576.  THEOREM.      The  lateral  faces  of  a  right  prism  are  rectangles. 
(Def.  569.) 

577.  THEOREM.     The  faces  and  bases  of  a  rectangular  parallele- 
piped are  rectangles.     (Def.  569.) 

578.  THEOREM.     All  the  faces  of  any  parallelepiped  are  paral- 
lelograms (?). 

579.  AXIOM.      A  polyhedron  cannot  have  fewer  than  four  faces. 

580.  AXIOM.     A  polyhedron  cannot  have  fewer  than  three  faces 
at  each  vertex. 


BOOK   VII 


295 


THEOREMS   AND   DEMONSTRATIONS 

581.   THEOREM.     The  sections  of  a  prism  made  by  parallel  planes 
cutting  all  the  lateral  edges  are  equal  polygons. 

Given  :  Prism  AB\  II  sections 
CF  and  C'F'. 

To  Prove: 

Polygon  (?>'=  polygon 

Proof  :  CD  is  II  to  C'D',  DE  is 
II  to  D'E',  etc.  (?)  (500). 

CDr,  DEr,  Erf,  etc.,  are  HJ  (?). 

.-.  CD  =  C'D',  DE=D'E',  EF 
=  E'F',  etc.  (?)  (130). 

Z  GCD  =  Z.  G'C'D',  Z.  CDE 
=^  C'D'E',  etc.  (?)  (515). 


.'.  polygon  OT=polygon  CfFf(?)  (159). 


J.E.D. 


582.   THEOREM.     The  opposite  faces  of  a  parallelepiped  are  equal 
and  parallel. 

Given:    (?).      To  Prove:  H—        G 

Face  AF  =  and  II  to  face  DG. 

Proof:  Faces  AF  and  DG 
are  £17  (?). 

AB=  DC,  AE  =  D#(130). 

AB  is  II  to  DC  and  AE  is  II  to 
DH  (?). 

.'.  /-  EAB  =  Z.  HDC  (?)  (515). 
.'.  face  AF  =  face  DG  (?)  (139). 
Also  face  AF  is  II  to  face  DG  (?)  (515). 


Q.E.D. 


Ex.  1.   All  right  sections  of  a  prism  are  equal. 

Ex.  2.    Any  section  of  a  parallelepiped  made  by  a  plane  cutting  two 
pairs  of  opposite  faces  is  a  parallelogram. 


296 


SOLID   GEOMETRY 


583.  THEOREM.  The  lateral 
area  of  a  prism  is  equal  to  the  prod- 
uct of  a  lateral  edge  by  the 
perimeter  of  a  right  section. 

Given:  Prism  EU1;  edge  = 
E)  right  section  AD. 

To  Prove:  Lateral  area  of 
EU1  =  E  x  perimeter  of  AD. 

Proof  :  AB  is  _L  to  EE1,  BG  is 
i  to  SS1,  etc.  (?)  (489). 

Area  O  RSf  =  E-  AB  (373). 
Area  O  STr  =  E  •  BC  (?). 
Area  O  TU'  =  ^  •  CD  (?). 

etc.          etc.         Adding, 

The  lateral  area  =  E-  (AB  +BC  +  CD  +  etc.)  (Ax.  2). 
=  jE?'  perimeter  of  rt.  sect.  (Ax.  6). 

Q.E.D. 

Ex.  1.  Any  section  of  a  parallelepiped  made  by  a  plane  parallel  to 
any  edge  is  a  parallelogram. 

Ex.  2.  The  sum  of  the  face  angles  at  all  the  vertices  of  any  parallele- 
piped is  equal  to  24  right  angles. 

Ex.  3.  The  sum  of  the  plane  angles  of  all  the  dihedral  angles  of  any 
parallelepiped  is  equal  to  12  right  angles. 

Proof :  Pass  three  planes  _L  to  three  intersecting  edges.  Prove  these 
sections  [U  whose  A  are  the  plane  angles  of  the  dihedral  angles,  etc. 

Ex.  4.   Enunciate  a  theorem  for  the  lateral  area  of  a  right  prism. 

—   Ex.  5.   Find  the  lateral  area  of  a  right  prism  whose  altitude  is  8  feet 
and  each  side  of  whose  triangular  base  is  5  feet.  I  V  0  fc 

*-   Ex.  6.   Find  the  total  area  of  a  regular  prism  whose  base  is  a  regular 
hexagon,  10  inches  on  a  side,  if  the  altitude  of  the  prism  is  15  inches.  °l 

>    Ex.  7.   Find  the  lateral  area  of  a  prism  whose  edge  is  12  inches  and 
whose  right  section  is  a  pentagon,  the  sides  of  which  are  3,  5,  6,  9,  and  L\  < 
11  inches. 


BOOK   VII 


297 


584.  THEOREM.  Two  prisms  are  equal  if  three  faces  including  a 
trihedral  angle  of  one  are  equal,  respectively,  to  three  faces  including 
a  trihedral  angle  of  the  other,  and  similarly  placed. 

Given  :  Prisms  AO  and  ArOf; 
face  AM  —  face  A'M* ';  face  AP 
=  face  A'P'-,  face  AD  =  face 
A'D'. 

To  Prove  :  Prism  AO  =  prism 


Proof :  The  three  face  A  at 
A  are  respectively  =  to  the  three  face  A  at  A1  (?)  (27). 

.-.  trih.  Z  A  =  trih.  Z  A1  (?)  (561). 

Superpose  prism  AO  upon  prism  A?o',  making  the  equal 
trihedral  A  A  and  A'  coincide. 

Face  AD  will  coincide  with  face  A'D',  face  A M  with  AfMf, 
face  ^1P  with  A'P'.     (They  are  =  by  hyp.) 

That  is,  point  L  will  fall  on  I/;  Jf  on  M'I  and  P  on  Pf. 
.-.  the  plane  iO  will  fall  upon  the  plane  L'O'  (?)  (493). 

Polygon  LO  —  polygon  L1  o'  (Ax.  1).     /.  these  bases  coin- 
cide (?). 

Similarly  face  BN  coincides  with  B'N'.,  CO  with  c'o'.    Etc. 
Consequently,  the  prisms  are  equal  (?)  (571).  Q.E.D. 

585.  THEOREM.     Two  right  prisms  are  equal  if  they  have  equal 
bases  and  equal  altitudes.     (Explain.) 

586.  THEOREM.     Two  truncated  prisms  are  equal  if  three  faces  in- 
cluding a  trihedral  angle  of  one  are  equal,  respectively,  to  three  faces 
including  a  trihedral  angle  of  the  other,  and  are   similarly  placed. 
(Explain.) 


298  SOLID  GEOMETRY 

587.  THEOREM.  An  oblique  prism  is  equivalent  to  a  right  prism 
whose  base  is  a  right  section  of  the  oblique  prism,  and  whose  altitude 
is  equal  to  the  lateral  edge  of  the  oblique  prism. 

p1  Or 


Given :    Oblique    prism 
right  prism  PN1  whose  base   is 
PN,  a  right  section  of  AC1,  and 
whose  altitude  PPf  =  edge  EE1. 

To  Prove  :   Oblique  prism  AC1 
=0=  right  prism  PNr. 

Proof:  Edge  EE'  =  PP'( Hyp.). 
Subtract    PE'    from    each,    and 

Likewise  AL  =  A'L',BM=B'M', 
CN=CrNr,  etc. 

(1)  Face  AC  =  face  A'Cf  (?)  (581). 

(2)  In  faces  AP   and  A'P', 

EP  =  EP',  AL  =  A'L'  (Ax.  2),  AE=A'E',  PL  =  P'L1  (?)(130). 
That  is,  face  AP  and  face  A'P'  are  mutually  equilateral. 
Also  Z  EAL  =  Z  E'A'L',  Z  PEA  =  Z  p'E*A\ 

ZA  T  "D        -    /     A     T     ~f^       s      ~IPT>T         -    /      Jv1'  T^  T  ^ 
_/\_  /y  r  —  £—  ^\_  JLt  Jr  «    •*-—  JL-L-Lj  —  -^—  Hf  -L     *J 

That  is,  face  AP  and  A'P'  are  mutually  equiangular. 
/.  face  AP  =  face  ArPf  (?)  (159). 

(3)  Similarly,  face  AM  =  face  A'M'. 

.'.  truncated  prism  A  N  —  truncated  prism  A'N'  (?)  (586). 
Now,       add,  solid  PC1  =  solid  PC1  (Iden.). 

Oblique  prism  AC'o=  right  prism  PN'  (Ax.  2).     Q.E.D. 


(?)  (98). 


Ex.  Prove,  in  the  figure  of  587,  that  truncated  prism  A  N  is  equal  to 
truncated  prism  A'N'  by  the  method  of  superposition. 

Proof:  Polygon  PN  =  polygon  P'N1  (V).  Superpose  solid  AN  upon 
solid  A'N'  so  that  base  PN  will  coincide  with  its  equal  P'N'.  Etc. 


BOOK   VII 


299 


588.  THEOREM.     The   plane  containing  two    diagonally  opposite 
edges  of  a  parallelepiped  divides  the  parallelepiped  into  two  equiva- 
lent triangular  prisms. 

Given :  Parallelepiped  EH 
and  plane  AG  containing  the 
opposite  edges  AE  and  CG. 

To  Prove:  Prism  ABC -F 
=c=  prism  ADC-H. 

Proof:  Pass  a  right  sec- 
tion BSTV  intersecting  the 
given  plane  in  RT. 

Face   AF  is  II  to  DG   (?). 

.-.  RS  is  II  to  VT  (?)  (500). 

Likewise,  RVis  II  to  ST  (?). 

.-.RSTVis  a  O  (?). 

/.  A  RST=  A  RVT  (?)  (132). 

Prism  ABC-F  =0=  a  right  prism  whose  base  is  R8T  and 
whose  altitude  =  EA  (?)  (587). 

Prism  ADC-H  =o=  a  right  prism  whose  base  is  RVT  and 
whose  altitude  =  EA  (?). 

But  these  imaginary  right  prisms  are  equal  (?)  (585). 

.'.  prism  ABC-F  =e=  prism  ADC-H  (Ax.  1).  Q.E.D. 

589.  THEOREM.     Two   rectangular   parallelepipeds   having    equal 
bases  are  to  each  other  as  their  altitudes. 

Given :  Rectangular  parallele- 
pipeds P  and  Q,  having  =  bases, 
and  their  altitudes  AB  and  CD, 
respectively. 

To  Prove  :   P  :  Q  =  AB  :  CD. 

Proof :  I .  If  the  altitudes  are 
commensurable. 

Consult  244,  302,  368,  539. 


B 

/ 

D 

/ 

0 

/ 

R 

/ 

'•/ 

:/ 

/ 

I 

300 


SOLID  GEOMETRY 


B 

D 
X 

C 

/ 

P 

/ 

R 

7\ 

^ 

- 

r 

- 

" 

/ 

" 

v 

II.    If  the  altitudes  are  incommensurable. 

There  does  not  exist  a  common  unit  (238). 

Suppose  AB  divided  into  equal 
parts.  Apply  one  of  these  as  a 
unit  of  measure  to  CD.  There 
will  be  a  remainder,  DX  (?). 

Pass  plane  XF,  through  X  and 

II  to  base.     Now,  —  =  —  (?). 
CY      CX 

Indefinitely  increase,  etc.,  as  in  244,  302,  368,  539. 

590.  THEOREM.  Two  rectangular  parallelepipeds  having  two  dimen- 
sions of  the  one  equal  respectively  to  two  dimensions  of  the  other, 
are  to  each  other  as  their  third  dimension. 

The  faces  having  the  sides  of  one  equal  to  the  sides  of  the  other,  re- 
spectively, may  be  considered  the  bases  and  the  third  dimensions  the 
altitudes.  Thus  this  statement  is  the  same  as  589. 

591.  THEOREM     Two    rectangular  parallelepipeds  having   equal 
altitudes  are  to  each  other  as  their  bases. 


/ 

R 

h 

/ 

/ 

I 

/ 

Given  :   Rect.  parallelepipeds  R  and  S,  having  the  same  alti- 
tude h ;  and  other  dimensions  a,  6,  and  c,  d,  respectively. 

To  Prove:  ?L=aLL^. 

S      c-d 

Proof :    Construct  a  third    rectangular    parallelepiped,  r, 
having  altitude  =  h,  another  dimension  =  a,  a  third  =  d. 

HandH<59°>     •••f-^C-Az.'S).  Q-E.D. 

^ 


BOOK  VII 


301 


592.  COR.     Two  rectangular  parallelepipeds  having  one  dimension 
in  common,  are  to  each  other  as  the  products  of  the  other  dimensions. 

593.  THEOREM.     Any  two  rectangular  parallelepipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


HI 


VL I 


T7 


b1 


Given :  Rectangular  parallelepipeds  L  and  J/,  whose  dimen- 
sions are  a,  6,  A,  and  a',  £>',  7i',  respectively. 

L        a-b-h 
To  Prove  :  — =  ~T~T7~~T7* 

Proof:  Construct  2V,  whose  dimensions  are  a,  5,  hf. 
Then,     |  =  p  (?)  (590),  and  J  =  ~  (?)  (592). 


I/      a-b-h      .  A       o-\ 
Multiplying,  x=af  -bf  -h'^  '' 


Q.E.D. 


594.     THEOREM.     The  volume  of   a  rectangular  parallelepiped  is 
equal  to  the  product  of  its  three  dimensions. 

Given:   (?)      To   Prove:  (?). 
Proof  :  Let  U  be  a  unit  of  vol. 


But_=  vol.  of  P  (?)  (571). 
.'.vol.  of  P  =  a  -  b  •  h  (Ax.  1). 


/ 

/ 

f=> 

/ 

-m-? 

w 

/• 

/ 

302 


SOLID   GEOMETRY 


595.  THEOREM.     The  volume  of   a  rectangular   parallelepiped  is 
equal  to  the  product  of  its  base  by  its  altitude.     (See  594.) 

596.  COR.     The  volume  of  a  cube  is  equal  to  the  cube  of  its  edge. 

597.  THEOREM.     The  volume  of  any  parallelepiped  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given:  Parallelepiped  #,  whose  base  =  B  and  alt.  =  h. 
To  Prove :  Volume  of  E  =  B  -  h. 

Proof :  Prolong  the  edge  AD  and  all  edges  II  to  AD.'  On 
the  prolongation  of  AD,  take  EF  =  AD.  Through  E  and  F 
pass  planes  EG  and  FH,  JL  to  EF,  forming  the  right  paral- 
lelepiped 8. 

Again,  prolong  FI  and  all  the  edges  II  to  FT.  On  the  pro- 
longation of  FI,  take  KL  =  FI.  Through  K  and  L  pass  planes 
KM  and  LN,  _Lto  KL,  forming  the  rectangular  parallelepiped  T. 

Consider  EG  the  base  of  s,  and  EF  its  altitude,  then  B  =c=  S 
(?)  (587).  Also  B  =0=  B '  (?)  (374). 

Consider  EP  the  base  of  S,  and  KM  the  base  of  T,  and 
KL  its  altitude,  then  S  =0=  T  (?)  (587).  Also  B'=  C  (?)(140). 

Hence,  R  =c=  r  (Ax.  1);  and  B  =0=  c  (Ax.  1);  and  altitude 
of  T  =  h  (?)  (524). 

But  volume  of  T  =  C  -  h  (?)  (595). 

/.volume  of  E  =  B  -  h  (Ax.  6).  Q.E.D. 


BOOK  VII  303 

598.  THEOREM.     Two  parallelepipeds  having  equal  altitudes  and 
equivalent  bases  are  equivalent  (Ax.  i). 

599.  THEOREM.     Two  parallelepipeds  having  equal  altitudes  are 
to  each  other  as  their  bases. 

Proof  :  Q  =  B  .  h  and  R  =  Bf  .  h  (?)  (597). 

.•.  by  dividing,  Q  =  —f  (Ax.  3).  Q.E.D. 

600.  THEOREM.     Two  parallelepipeds  having  equivalent  bases  are 
to  each  other  as  their  altitudes  (?). 

601.  THEOREM.     Any  two  parallelepipeds  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes  (?). 

602.  THEOREM.     The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Given  :  Triangular  prism 
ACD-F;  base  =  £;  alt.  =A. 

To  Prove: 
Volume  of   ACD-F=B-h. 

Proof:  Construct  parallele- 
piped AS  having  as  three  of 
its  lateral  edges  AE,  CF,  DG. 


Hence,  |  volume  of  AS  —  ^  ACRD  •  h  (Ax.  3). 
But  i  volume  of  AS=  volume  of  prism  ACD-F  (?)  (588) 
andj  ACRD  =  B  (?)(132). 

.'.  volume  of  ACD-F  =  B  -  h  (Ax.  6).  Q.E.D. 


Ex.  1.  Which  rectangular  parallelepiped  contains  the  greater  volume, 
one  whose  edges  are  5,  7,  9,  or  one  whose  edges  are  4,  6,  13  ? 

Ex.  2.  The  base  of  a  prism  is  a  right  triangle  whose  legs  are  8  and 
12,  and  the  altitude  of  the  prism  is  20.  Find  its  volume. 


304  SOLID  GEOMETRY 

603.   THEOREM.     The  volume  of  any  prism  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Given  :  Prism  AD ;  base  =  B  ; 
altitude  =  h. 

To  Prove:    Vol.  of  AD=B-h. 

Proof :  Through  any  lateral 
edge,  AC,  and  other  lateral  edges 
not  adjoining  AC,  pass^  planes 
cutting  the  prism  into  triangular 
prisms  I,  II,  III,  having  bases 
E,  s,  T,  respectively. 

Vol.  of  prism      I  =  E  -  h 


Vol.  of  prism    II  =  8  -  h  f  (?)  (602). 

Vol.  of  prism  III  =  T-  h  J  Adding, 

Vol.  of  prism  AD  =  (#  -f  s  +  T)h  =  B  •  h  (Ax.  2).         Q.E.D. 

604.  THEOREM.     Two  prisms  having  equal  altitudes  and  equiva- 
lent bases  are  equivalent. 

605.  THEOREM.     Two  prisms  having  equal  altitudes  are  to  each 
other  as  their  bases. 

606.  THEOREM.     Two  prisms  having  equivalent  bases  are  to  each 
other  as  their  altitudes. 

607.  THEOREM.     Any  two  prisms  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes. 

ORIGINAL  EXERCISES 

1.  How  many  faces  has  a  parallelepiped?    Edges?    Vertices?    How 
many  faces  has  a  hexagonal  prism?     Edges?     Vertices? 

2.  Every  lateral  face  of  a  prism  is  parallel  to  the  lateral  edges  not 
in  that  face. 

3.  Every  lateral  edge  of  a  prism  is  parallel  to  the  faces  that  do  not 
contain  it. 

4.  Every  plane  containing  one  and  only  one  lateral  edge  of  a  prism 
is  parallel  to  all  the  other  lateral  edges. 


BOOK   VII 


305 


5.  Any  lateral  face  of  a  prism  is  less  than  the  sum  of  the  other 
lateral  faces.     [Use  fig.  of  583.] 

6.  The  diagonals  of  a  rectangular  parallelepiped  are  equal. 
Proof:   Pass  the  plane  A  CGE.    This  is  a  rectangle  (?),  etc. 


7.  The  four  diagonals  of  a  parallelepiped  bisect 
each  other. 

[First  prove  that  one  pair  bisect  each  other,  thus 
prove  that  any  pair  bisect  each  other,  etc.] 

8.  Two   triangular  prisms    are    equal    if    their 
lateral  faces  are  equal  each  to  each. 


B 


9.    Any  prism  is  equivalent  to  the  parallelepiped  having  the  same 
altitude  and  an  equivalent  base. 

10.  The  square  of  the  diagonal  of  a  rectangular 
parallelepiped  is  equal  to  the  sum  of  the  squares  of 
its  three  dimensions. 

To  Prove:   ZC2  =  AE'2  +  ED2  +  DC*. 
Proof :  A  D  is  the  hypotenuse  of  rt.  A  A  ED,  and 
AC,  of  rt.&ACD. 

11.  The  diagonal  of  a  cube  is  equal  to  the  edge  multiplied  by  \/3. 

12.  The  volume  of  a  triangular  prism  is  equal  to  half  the  product  of 
the  area  of  any  lateral  face  by  the  perpendicular  drawn  to  that  face  from 
any  point  in  the  opposite  edge.    [Use  the  fig.  of  602.] 

13.  Every  section  of  a  prism  made  by  a  plane 
parallel  to  a  lateral  edge  is  a  parallelogram. 

To  Prove :   LMRN  a  O.    Proof :   LM  is  \\  to  NR 
(?).   LN  and  MR  are  each  \\  to  any  edge.  (Explain.) 

14.  Every  polyhedron  has  an  even  number  of  face 
angles. 

Proof:  Consider  the  faces  as  separate  polygons, 
sides  of  these  polygons  =  double  the  number  of  edges  of  the  polyhedron. 
(Explain.)  But  the  number  of  sides  of  these  polygons  =  the  number  of 
their  angles,  that  is,  the  number  of  face  angles.  .-.  the  number  of  face 
angles  =  double  the  number  of  edges  =  an  even  number  (?). 

15.  There  is  no  polyhedron  having  fewer  than  6  edges. 


The  number  or 


306  SOLID   GEOMETRY 

16.  A  room  is  7  m.  long,  5  m.  wide,  3  m.  high.     Find  its  contents 
and  its  total  area. 

17.  Find  the  volume,  lateral  area,  and  total  area  of  an  8-in.  cube. 

18.  A  right  prism  whose  height  is  12  ft.  has  for  its  base  a  right 
triangle  whose  legs  are  6  ft.  and  8  ft.     Find  the  volume,  lateral  area, 
and  total  area  of  the  prism. 

19.  Find  the  altitude  of  a  rectangular  parallelepiped  whose  base  is 
21  in.  x  30  in.,  equivalent  to  a  rectangular  parallelepiped  whose  dimen- 
sions are  27  in.  x  28  in.  x  35  in. 

;    20.    A  cube  and  a  rectangular  parallelepiped  whose  edges  are  6,  16, 

and  18,  have  the  same  volumes.     Find  the  edge  of  the  cube,     f  i/ 

^j    21.    Find  the  volume  of  a  rectangular  parallelepiped  whose  total  area 
is  620  and  whose  base  is  14  x  9.   V  "^  ^  »  ^(  "^  \  A  0  $  £/w ,  vwUAfl^ 

22.  How  many  bricks  each  8  x  2|  x  2  in.  will  be  required  to  build  a 
wall  22  x  3  x  2  ft.  (not  allowing  for  mortar)  ?  <T  /  y  y, 

23.  If  a  triangular  prism  is  20  in.  high  and  each  side  of  its  base  is 
8  in.,  how  many  cubic  inches  does  it  contain?   -*,  -^  Q  \f^ 

^^    24.   Find  the  lateral  area,  total  area,  and  volume  of  a  regular  hexagonahJ 
prism  each  side  of  whose  base  is  10  and  whose  altitude  is  15.  |L  "  fOO 

s  25.   A  box  is  12  x  9  x  8  in.     What  is  the  length  of  its  diagonal?    ,  r, 

^  "^C-. 

26.   Each  edge  of  a  cube  is  8  in.     Find  its  diagonal. 

^.  27.   The  diagonal  of  a  cube  is  10  A/3.    Find  its  edge,  volume,  total  area. 

^-  28.  A  trench  is  180  ft.  long  and  12  ft.  deep,  7  ft.  wide  at  the  top  and 
4  ft.  at  the  bottom.  How  many  cubic  yards  of  earth  have  been  removed  ?  ut 

,  29.  A  metallic  tank,  open  at  the  top,  is  made  of  iron  2  in.  thick ; 
the  internal  dimensions  of  the  tank  are,  4  ft.  8  in.  long,  3  ft.  6  in.  wide, 
4  ft.  4  in.  deep.  Find  the  weight  of  the  tank  if  empty ;  if  full  of  water. 
[Water  weighs  62|  Ib.  to  the  cu.  ft.  and  iron  is  7.2  times  as  heavy  as  water.] 

/*  30.  The  base  of  a  right  parallelepiped  is  a  rhombus  whose  sides  are 
each  25,  and  the  shorter  diagonal  is  14.  The  height  of  the  parallelepiped 
is  40.  Find  its  volume  and  total  surface.  Au  \  vq  ^ 

~-  31.   If  the  diagonal  of  a  cube  is  12  ft.,  find  its  surface.  /f£X  U 

32.  If  the  total  surface  of  a  cube  is  1  sq.  yd.,  find  its  volume  in  cu.  ft. 

33.  A  right  prism  whose  altitude  is  25  has  for  its  base  a  triangle  whose 
sides  are  11,  13,  20.     Find  its  lateral  area,  total  area,  and  volume. 


BOOK  VII  307 

PYRAMIDS 

608.  A  pyramid  is  a  polyhedron,  one  of  whose  faces  is  a 
polygon  and  whose  other  faces  are  all  triangles  having  a 
common  vertex. 

The  lateral  faces  of  a  pyramid  are  the  triangles. 

The  lateral  edges  of  a  pyramid  are  the  intersections  of  the 
lateral  faces.  The  vertex  of  a  pyramid  is  the  common  vertex 
of  all  the  lateral  faces. 

The  base  of  a  pyramid  is  the  face  opposite  the  vertex. 

The  lateral  area  of  a  pyramid  is  the  sum  of  the  areas  of 
the  lateral  faces.  The  total  area  of  a  pyramid  is  the  sum 
of  the  lateral  area  and  the  area  of  the  base. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance 
from  the  vertex  to  the  plane  of  the  base. 

A  triangular  pyramid  is  a  pyramid  whose  base  is  a  tri- 
angle. It  is  called  also  a  tetrahedron.  (See  566.) 


PYRAMIDS  REGULAR         TRUNCATED    FRUSTUM  OP 

PYRAMIDS         PYRAMID      A  PYRAMID 


609.  A  regular  pyramid  is  a  pyramid  whose  base  is  a  regu- 
lar polygon  and  whose  altitude,  from  the  vertex,  meets  the 
base  at  its  center. 

The  slant  height  of  a  regular  pyramid  is  the  line  drawn 
in  a  lateral  face,  from  the  vertex  perpendicular  to  the  base 
of  the  triangular  face.  It  is  the  altitude  of  any  lateral  face. 


308  SOLID   GEOMETRY 

610.  The  frustum  of  a  pyramid  is  the  part  of  a  pyramid 
included  between  the  base  and  a  plane  parallel  to  the  base. 

The  altitude  of  a  frustum  of  a  pyramid  is  the  perpendicu- 
lar distance  between  the  planes  of  its  bases. 

The  slant  height  of  the  frustum  of  a  regular  pyramid  is 
the  perpendicular  distance,  in  a  face,  between  the  bases  of 
that  face. 

A  truncated  pyramid  is  the  part  of  a  pyramid  included 
between  the  base  and  a  plane  cutting  all  the  lateral  edges. 

PRELIMINARY   THEOREMS 

611.  THEOREM.     The  lateral  edges  of  a  regular  pyramid  are  all 
equal.     (See  520,  II.) 

612.  COR.     The  lateral  faces  of  a  regular  pyramid  are  equal  isos- 
celes triangles. 

613.  COR.   The  lateral  edges  of  the  frustum  of  a  regular  pyramid 
are  all  equal.     (Ax.  2.) 

614.  THEOREM.     The  lateral  faces  of  the  frustum   of    a  regular 
pyramid  are  equal  isosceles  trapezoids.    (See  500.) 

615.  THEOREM.     The  lateral  faces  of  the  frustum  of  any  pyramid 
are  trapezoids.     (?). 

616.  THEOREM.     The  slant  height  of  a  regular  pyramid  is  the 
same  length  in  all  the  lateral  faces. 


Ex.  1.  Prove  that  the  bases  of  any  frustum  of  a  pyramid  are 
mutually  equiangular. 

Ex.  2.  The  foot  of  the  altitude  of  a  regular  pyramid  drawn  from  the 
vertex,  coincides  with  the  center  of  the  circles  inscribed  in,  and  circum- 
scribed about,  the  base. 

Ex.  3.  The  sum  of  the  medians  of  the  lateral  faces  of  the  frustum  of 
a  trapezoid  is  equal  to  half  the  sum  of  the  perimeters  of  the  bases. 


BOOK  VII 


309 


THEOREMS   AND   DEMONSTRATIONS 

617.   THEOREM.     The  lateral  area  of  a  regular  pyramid  is  equal 
to  half  the  product  of  the  perimeter  of 
the  base  by  the  slant  height. 

Given :  Regular  pyramid  o- 
ABCDE;  lat.  area  =  i;  perimeter 
of  base  =  P ;  slant  height  OH  =  8. 

To  Prove  :  L  =  J  P  •  s. 

Proof : 

Area  AAOB=  \AB- 

Area  A  BOC=  \BC- 

etc.  etc.  Adding, 

Lateral  area  =  J  AB  •  s  +  |-  BC  -  s  +  etc.  (Ax.  2). 

That  is,     i  =  ^(AB  -\-BC+  etc.)  .  s,  or, 

L  =  |  P  •  s  (Ax.  6).  Q.E.D. 


618.  THEOREM.  The  lateral  area  of  the 
frustum  of  a  regular  pyramid  is  equal  to  half 
the  sum  of  the  perimeters  of  the  bases  mul- 
tiplied by  the  slant  height. 

Given:   (?). 
To  Prove:  i 


Proof :  Area  trapezoid  Cl  =  J(CD  +  Hi)  -  s  (?). 
Area  trapezoid  BH  =  ^(JBC  +  GB)  -  s  (?). 
Area  trapezoid  AG  =  ^(AB+  FG)  -  s  (?).  Etc. 


Adding,  lateral  area  Al= 


s.  (Explain.)  Q.E.D. 


Ex.  1.  The  slant  height  of  a  regular  pyramid  whose  base  is  a  square, 
of  which  each  side  is  8  ft.,  is  15  ft.  Find  the  lateral  area;  the  total  area. 

Ex.  2.  A  regular  pyramid  stands  on  a  hexagonal  base  16  in.  on  a 
side,  and  the  slant  height  is  2  ft..  Find  the  lateral  and  total  areas. 


310  SOLID  GEOMETRY 

619.   THEOREM.     If  a  pyramid  is  cut  by  a  plane  parallel  to  the 
base: 

I.   The  lateral  edges  and  altitude  are  divided  proportionally. 

II.   The  section  is  a  polygon  simi-  O 

lar  to  the  base. 

Given:  Pyr.  O-ABCDE-,    plane 
FI II  to  the  base ;  altitude  =  OL. 

To  Prove : 

j     OF__OG__OH_       _  OM 
OA  ~~  OB       OC  OL 

II.    Section    FI   is    similar    to 
the  base  AD. 


Proof:    I.    Imagine  a  plane  through  o  II  to  plane  AD. 
This  plane  is  -L  to  OL  (512),  and  II  to  FI  (?)  (505). 
...of=o«  =  off=...  =  o£ 

OA       OB       OC  OL   ^ 

II.    FG  is  II  to  AB,  GH  is  II  to  BC,  etc.  (?)  (500). 
.'.  Z.FGH  =  Z-  ABC',   Z  GUI  =  Z  BCD  ;  etc.  (?)  (515). 

That  is,  the  polygons  are  mutually  equiangular.     Also, 
A  OFG  is  similar  to  A  OAB  ;  A  OGH  to  A  OBC  ;   etc.  (?)  (316). 

FG       fOG\       GH       fOH\       HI 

.-. =    —    =  —  =   -  -]  =  —  =  etc.  (?)  (323,  3). 

AB       \OBj       BC       \OCj       CD 

.'.  section  FI  is  similar  to  base  AD  (?)  (312).  Q.E.D. 


Ex.  1.  The  bases  of  the  frustum  of  a  regular  pyramid  are  equilateral 
triangles  whose  sides  are  12  in.  and  20  in.,  respectively.  The  slant 
height  is  40  in.  Find  the  lateral  area;  the  total  area. 

Ex.  2.  The  bases  of  frustum  of  a  regular  pyramid  are  regular  hexa- 
gons whose  sides  are  8  and  18,  respectively.  The  slant  height  is  25. 
Find  the  lateral  area  and  total  area. 


BOOK  VII  311 


620.  THEOREM.  If  two  pyramids  have  equal  altitudes  and  equiva- 
lent bases,  sections  made  by  planes  parallel  to  the  bases  and  at  equal 
distances  from  the  vertices  are  equivalent. 


Given  :   Pyramids  O-ABCDE  and  o'-PQRS  ;  alt.  OL  =  O'L'  ; 
base  AD  =0=  base  PR  ;  sections  FI  and  TV  II  to  bases  ;   OM  =  O1  M'  . 
To  Prove  :  Section  FI  =0=  section  TV. 

Proof  :'  FG  is  II  to  AB  ;   TU  is  II  to  PQ,  etc.  (?)  (500). 
A  OFG  and  OAB  are  similar,  also  A  o'TU  and  o'PQ  (?)(316). 

.-.  —  =  —  =  —  (?)  (323,  3  and  619,  I). 
AB       OA       OL  ^ 

And  IE=<?T  =  ^(?).  But  OM=  O'M'  and  OL  =  O'L'  (Hyp.). 
PQ       O'P       L'O'^ 


Now  section  FI  is  similar  to  base  AD,  and  section  TV  is 
similar  to  base  PR  (?)  (619,  II). 


section  FI      FG2       -    section  TV     TU2 


. 

base  ^4D       ^#  base  Ptf 

Hence,  section  ^=section  rTr(Ax.  1). 
base  AD         base  P# 

But  base  AD  =c=  base  P.R  (Hyp.). 

,-.  section  F/O  section  TV  (?)  (Ax.  3)  Q.E.D. 


\\ 


312 


SOLID   GEOMETRY 


621.  If  a  plane  be  passed  parallel  to  the  base  of  a  pyramid,  intersect- 
ing all  the  lateral  edges,  and  upon  the  section  thus  formed,  as  a  base,  a 
prism  be  constructed  wholly  inside  the  pyramid,  but  having  one  lateral 
edge  in  a  lateral  edge  of  the  pyramid,  this  prism  is  called  an  inscribed 
prism.  (See  622,  fig.  1.) 

If  upon  this  section  as  a  base  a  prism  be  constructed  partly  outside 
the  pyramid,  having  one  lateral  edge  in  one  of  the  lateral  edges  of  the 
pyramid,  this  prism  is  called  a  circumscribed  prism.  (See  622,  fig.  2.) 

X""*"  ""V 

•V622.  THEOREM.  The  volume  of  a  triangular  pyramid  is  the  limit 
of  the  sum  of  the  volumes  of  a  series  of  inscribed  or  circumscribed 
prisms,  having  equal  altitudes,  if  the  number  of  prisms  is  indefinitely 
increased. 


Given :  Triangular  pyramid  O-ABC,  having  a  series  of 
prisms  inscribed  in  it,  and  another  series  circumscribed  about 
it,  all  the  prisms  having  equal  altitudes. 

To  Prove  :  O-ABC  is  the  limit  of  the  sum  of  each  series 
of  prisms  as  their  number  is  indefinitely  increased. 

Proof  :  Denote  the  volume  of  the  pyramid  by  F,  the  sum 
of  the  volumes  of  the  series  of  inscribed  prisms  by  S&  and  the 
sum  of  the  volumes  of  the  series  of  circumscribed  prisms  by  8C. 

The  uppermost  circumscribed  prism  =0=  the  uppermost  in- 
scribed prism  (604). 

The  second  pair  of  prisms  also  are  equivalent  (?). 


BOOK  VII  313 

And  so  on,  until  the  last  circumscribed  prism,  D-ABC, 
remains,  for  which  there  is  no  equivalent  inscribed  prism. 

Hence,  it  is  evident  that  Sc—  S^D  —  ABC,  the  lowest  prism. 

Now  by  indefinitely  increasing  the  number  of  the  prisms, 
the  altitude  of  D—ABC  will  become  indefinitely  small,  and 
hence  the  volume  of  D—ABC  will  approach  zero  as  a  limit. 

The  altitude  can  never  actually  equal  zero,  nor  can  the  vol- 
ume equal  zero.     Hence,  Sc—  8t  can  be  made  less  than  any 
mentionable  quantity,  but  cannot  equal  zero. 
Now    sc  =  sc  V  <  Sc  (Ax.  5) 

and     V  >  8j  (Ax.  5).  and     st  =  st 

.-. sc—  v <  se  —  Si  (Ax.  9).        .*.  F  —  st  <  sc—  8t  (Ax.  7). 

That  is,  sc—  V  and  F—  S<  are  each  less  than  sc—  S&  which 
itself  approaches  zero. 

Hence,  8e  —  F  approaches  zero  and  F  —  S{  approaches  zero. 

.-.  Sc  approaches  F  as  a  limit,  and  st  approaches  F  as  a 
limit  (?)  (240).  (See  note  on  p.  223.)  Q.E.D. 


Ex.  1.  If  a  plane  is  passed  parallel  to  the  base  of  a  pyramid,  cutting 
the  lateral  edges,  the  section  is  to  the  base  as  the  square  of  the  altitude 
of  the  pyramid  cut  away  by  this  plane  is  to  the  square  of  the  altitude  of 
the  original  pyramid.  (See  proof  of  620.) 

Ex.  2.  If  two  pyramids  have  equal  altitudes  and  are  cut  by  planes 
parallel  to  the  bases  and  at  equal  distances  from  the  vertices,  the  sections 
formed  will  be  to  each  other  as  the  bases  of  the  pyramids. 

Ex.  3.  In  the  figure  of  622  prove  the  planes  of  the  faces  of  the  prisms, 
that  are  opposite  OC,  are  parallel  to  OC  and  to  AB. 

Ex.  4.   State  the  theorems  leading  up  to  the  theorem  of  588. 

Ex.  5.   State  the  theorems  leading  up  to  the  theorem  of  597. 

Ex.  6.   State  the  theorems  leading  up  to  the  theorem  of  603. 

Ex.  7.  The  base  of  a  pyramid  is  180  sq.  in.  and  its  altitude  is  15  in. 
What  is  the  area  of  the  section  made  by  a  plane  parallel  to  the  base, 
and  5  in.  from  the  vertex  ?  Iff  I  £  •»•  I  #  0  I  Jt  >  7^-  T*0 

Ex.  8.   The  base  of   a  pyramid   is    200   sq.  in.,  and  its   altitude  is 
12  in.     At  what  distance  from  the  vertex  must  a  plane  be  passed  so 
that  the  section  shall  contain  half  the  area  of  the  base  ? 
{  V     '«  7U*  ¥.    V  0  0   ,'    f  O*  O 


314 


SOLID   GEOMETRY 


623.    THEOREM.     Two  triangular  pyramids  having  equal  altitudes 
and  equivalent  bases  are  equivalent. 

o  or 


Given:  Triangular  pyramids  O-ABC  and  Of-AfBrCr  having 
equal  altitudes  and  base  ABC  =0=  base  A'B'C!. 

To  Prove:  O-ABC  =c=  O'-A'B'C'. 

Proof :  Divide  the  altitude  of  each  pyramid  into  any  num- 
ber of  equal  parts.  Through  these  points  of  division  pass 
planes  II  to  the  bases,  forming  triangular  sections. 

Upon  these  sections  as  bases  construct  inscribed  prisms. 

Denote  the  volumes  of  the  pyramids  by  V  and  F;,  and  the 
sums  of  the  volumes  of  these  series  of  prisms  by  8  and  Sr. 

The  corresponding  sections  are  =0=  (?)  (620). 

.*..  the  corresponding  prisms  are  =c=  (?)  (604). 

Hence,  s=sf  (Ax.  2). 

By  indefinitely  increasing  the  number  of  equal  parts  into 
which  the  altitudes  are  divided,  the  number  of  prisms  be- 
comes indefinitely  great. 

.*.  8  approaches  V  as  a  limit,  (?)  (622), 
and  Sr  approaches  V1  as  a  limit   (?). 

.-.  v  -*  v'  (?)  (242).      That  is,  O-ABC  =c=  Or-AfBfCf. 

Q.B.D. 

NOTE.  As  in  plane  geometry,  A  ABC  is  the  same  as  A  BA C,  so  in 
solid  geometry  the  pyramid  O-ABC  is  the  same  as  the  pyramid  A -B CO 
or  B-A  CO  or  C-ABO. 


BOOK  VII 


315 


624.    THEOREM.     The  volume  of  a  triangular  pyramid  is  equal  to 
one  third  the  product  of  its  base  by  its  altitude. 

Given:  Triangular  pyramid 
O-AMC,  whose  base  =  B  and  alti- 
tude =  h. 

To  Prove  : 
Volume  O-AMC  =  ^  B  -  h. 

Proof:    Construct  a  prism  AMC- 
DOE,  having  AMC  as  its  base,  and       , 
OM  as  one  of  its  lateral  edges. 

Pass  a  plane  through  DO  and 
OC,  cutting  the  face  AE  in  line  CD. 

The  prism  is  now  divided  into  three  triangular  pyramids. 

In  pyramids  O-AMC  and  C-ODE,  the  altitudes  are=(?)(524). 

The  bases  AMC  and  ODE  are  equal  (?)  (568). 

.-.  pyramid  O-AMC  =c=  pyramid  C-ODE  (?)  (623). 
In  the  pyramids  C-AMO  and  C-AOD,  the  altitudes  are  the 
same  line  from  c  _L  to  plane  DM  (?)  (507). 
The  bases  AMO  and  AOD  are  =  (?)  (132). 
.*.  pyramid  C-AMO  =0=  pyramid  C-AOD  (?)  (623). 
Hence,  O-AMC  =0=  C-ODE  =0=  C-AOD  (Ax.  1). 
That  is,  O-AMC  =  ^  the  prism. 
But  the  volume  of  the  prism  =  B  •  h  (?). 
/.  volume  of  pyramid  o-AMC=  J  B-  h  (Ax.  6).      Q.E.D. 


Ex.  1.   In  the  figure  of  624,  prove  pyramid  0-A  CD  -  0-CDE. 
„__    Ex.  2.    The    area  of  the  base  of  a  triangular  pyramid  is  30  sq.  in., 
and  its  altitude  is  20  in.      Find  the  volume.     Find  the  volume  of  the 
prism  having  the  same  base  and  altitude.  j*4M^  <2^.  '*++,  ^    (ffotf  «xc, 

Ex.  3.    A  pyramid  whose  base  is  b  and  altitude  is  h  is  equivalent  to 
another  pyramid  whose  base  is  d  and  altitude  is  x.     Find  x. 

Ex.  4.   If  in  the  figure  of  602,  a  plane  is  passed  through  E  and  CD, 
what  part  of  the  whole  parallelepiped  is  the  pyramid  E-A  CD? 


316 


SOLID   GEOMETRY 


. 


625.   THEOREM.     The  volume  of  any  pyramid  is  equal  to  one  third 
the  product  of  its  base  by  its  altitude. 

Given:   Pyramid  o-CDEFG,  whose 
base  =  B  and  altitude  =  h. 

To  Prove : 

Volume  of  O-CDEFG  ==  J  B  -  h. 

Proof :    Through  any  lateral  edge, 
OC,  and  lateral  edges  not  adjoining 
C,  pass  planes  dividing  the  pyra- 
mid into  triangular  pyramids. 
Vol.  of  0-CDE=±CDE-h] 
Vol.  of  O-CEF=ICEF-  h 


Vol.  of 


=  ±CFG-h 


Adding, 


Vol.  of  o-CDEFG=^B-h  (Ax.  2  and  Ax.  4). 


Q.E.D. 


626.  THEOREM.     Any  two  pyramids  having  equal  altitudes  and 
equivalent  bases  are  equivalent.     (Ax.  1.) 

627.  THEOREM.     Two  pyramids  having  equal  altitudes  are  to  each 
other  as  their  bases.     (Prove.) 

628.  THEOREM.     Two  pyramids   having  equivalent  bases  are  to 
each  other  as  their  altitudes.     (Prove.) 

629.  THEOREM.     Any  two  pyramids  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes.     (Prove.) 

{630.  THEOREM.  The  volume  of  the  frustum  of  a  triangular  pyra- 
mid is  equal  to  one  third  the  altitude  multiplied  by  the  sum  of  the 
lower  base,  the  upper  base,  and  a  mean  proportional  between  the  bases 
of  the  frustum. 

Given :    The  frustum  BD  of  a  triangular  pyramid  whose 
lower  base  =  B  ;  upper  base=  b ;  altitude  =  h. 


To  Prove:    Volume  of  BD  =  $h  [B  +  b  +  V 


\\ 


BOOK  VII 


317 


Proof  :  Pass  a  plane  through 
edge  CE  and  vertex  s,  and 
another  through  edge  RS  and 
vertex  E,  dividing  the  frustum 
into  three  triangular  pyramids, 
S-CDE,  E-RST,  E-CRS. 


II. 

III.   We    shall    now    prove 
E-CRS  =     h-  VjT~b. 


E-CSD  _  A  CSD 
E-CRS      A  CRS 


(?)  (627). 


= 

RS       ^ 


T  .-,        . 
Likewise, 


E-CRS      US 

S-CER  A 
-  —  =- 
S-ERT  A 


=        (Ax.  1). 

A 


XON         -, 
(?)  and 


A  ERT      ET 


S-CER  _  CE 
~  RT 


But  A  CDE  and  RST  are  similar  (?)  (619,  II). 

CD_CE  ,?. 
'  RS      RT 

TT  E-CSD      S-CER  or  E-CRS   ,  A'       1N 

Hence,  -        —  =  —  -  (Ax.  1). 


E-CRS 


S-ERT 


—  ~h  •  7?        7?1— 

That  is,  ^        -  =  -!—-  (Substituting  from  I  and  II). 
E—CRS        -k  fl  •  0 


/.  E-CRS 


(?)  (299). 


.*.  volume  of  the  frustum  =  J  h  \_B  +  6  +  V^  •  b~\  (Ax.  2). 

Q.E.D. 

NOTE.     Theorem  630  is  sometimes  stated  thus  : 

The  frustum  of  a  triangular  pyramid  is  equivalent  to  the  sum  of 
three  pyramids  whose  altitudes  are  the  same  as  the  altitude  of  the  frus- 
tum and  whose  bases  are  the  lower  base,  the  upper  base,  and  a  mean 
proportional  between  the  bases  of  the  frustum. 


318  SOLID  GEOMETRY 

631.  THEOREM.  The  volume  of  the  frustum  of  any  pyramid  is 
equal  to  one  third  the  altitude  multiplied  by  the  sum  of  the  lower 
base,  the  upper  base,  and  a  mean  proportional  between  the  bases  of 
the  frustum. 

Given:  Pyr.  O-ADEFG', 
frustum  A'F,  whose  lower 
base  =  J5,  upper  base  =  6, 
altitude  =  h. 

To  Prove:  Vol.  of  frus- 
tum =  ^h[B  +  b  + 


Proof :  Construct  a  A  QRS  =0=  polygon  AF  (by  409). 

Upon  A  QRS  as  a  base,  construct  a  pyramid  whose  altitude 
=  the  altitude  of  O-ADEFG.  Pass  a  plane  Q'R'S'  II  to  QRS  and 
at  a  distance  from  QRS  =  h. 


Vol.  of  Q'JR  =  lA[A  QRS  +  A  Q'R'S' +  VA  QRS  •  A  Q'R's1]  (630). 
The  alt.  of  P-Q'R'S'  =  alt.  of  O-A'D'E'F'G'  (?)  (Ax.  2). 

Also,  QRS^B  (Const.);  and  Q'R'S' ^b  (?)  (620). 
.'.  vol.  of  O-ADEFG  =  vol.  of  P-QRS  (?)  (626),  and 

vol.  of  0-A'D'E'F'G'  =  vol.  of  P-Q'R'S'  (?).          Subtracting, 

vol.  of  frustum  A'F=  vol.  of  frustum  Q'R  (Ax.  2). 

Vol.  of  frustum  A'F=  J  h[B+b  +  VB  •  b]  (Ax.  6).       Q.E.D. 


Ex.  1.  Find  the  volume  of  a  pyramid  whose  altitude  is  18  in.  and 
whose  base  is  10  in.  square.  bU*A/vx.  '^~ 

Ex.  2.  Find  the  volume  of  the  frustum  of  a  pyramid  whose  altitude 
is  20  and  the  areas  of  whose  bases  are  18  and  32.  7,-d,  ^  c^J, 

Ex.  3.  The  bases  of  the  frustum  of J a  pyramid  are  regular  hexagons 
whose  sides  are  10  in.  and  6  in.,  respectively.  The  altitude  of  the  frus- 
tum is  2  ft.  Find  its  volume.  %  %  ^  v  ^__^ 

* 


BOOK  VII 


319 


632.  THEOREM.  A  truncated  triangular  prism  is  equivalent  to 
three  triangular  pyramids  whose  bases  are  the  base  of  the  prism  and 
whose  vertices  are  the  three  vertices  of  the  face  opposite  the  base  (the 
inclined  section). 


Given :  The  truncated  triangular  prism  ABC-RST,  whose 
base  is  ABC  and  whose  opposite  vertices  are  R,  S,  T.  Let  it 
be  divided  by  the  planes  ACS,  ABT,  BCR. 

To  Prove  :    ABC-RST  0=  R-ABC  +  S-ABC  +  T-ABC  (III). 

Proof :  In  Fig.  I,  S-ABCis  obviously  one  of  these  pyramids. 
In  Fig.  II,     A-CST  =0=  A-BCT  (?)  (626). 
That  is,  A-C8T  =0=  T-ABC. 

In  Fig.  Ill,   T-ARS  =0=  T-ABR  (?)  (626). 
T-ABR  =0=  C-ABR  (?)  (626). 
.*.  T-ARS  =c=  R-ABC  (Ax.  1). 
Now,  ABC-RST  =c=  T-ARS  +  S-ABC  +  A-CST  (Ax.  4). 


Hence,    ABC-RST 


R-ABC  +  S-ABC  +  T-AB C  (Ax.  6). 

Q.E.D. 


633.  COR.     The  volume  of  a  truncated  triangular 
prism  is  equal  to  the  product  of  the  base  by  one  third 
the  sum  of  the  three  altitudes  drawn  to  the  base  from 
the  three  vertices  opposite  the  base, 

634.  COR.     The  volume  of  a  truncated  right  tri- 
angular prism  is  equal  to  the  product  of  the  base  by 
one  third  the  sum  of  its  lateral  edges. 


320 


SOLID   GEOMETKY 


635.  THEOREM.  The  volume  of  any  truncated 
triangular  prism  is  equal  to  the  product  of  its  right 
section  by  one  third  the  sum  of  its  lateral  edges. 

Proof:  The  right  section  divides  the  solid 
into  two  truncated  right  prisms. 

Hence,  volume  =  right  section  x  ^  sum  of 
lateral  edges  (634).  Q.E.D. 


636.  THEOREM.  Two  triangular  pyramids  (tetrahedrons)  having 
a  trihedral  angle  of  one  equal  to  a  trihedral  angle  of  the  other  are 
to  each  other  as  the  products  of  the  three  edges  including  the 
equal  trihedral  angles. 


Given:  Triangular  pyra- 
mids S-ABC,  S-PQR-,  having 
the  trih.  A  at  8  equal ;  and 
their  volumes  F  and  V1. 

To  Prove :  —,  =  ° 


V1       SP-SQ-SR 

Proof:    Place    the    pyra- 
mids so  that  the  equal  trihedral  A  coincide.     Draw  the  alti- 
tudes AX  and  PY  and  the  projection  SXY  in  plane  SQR. 

A  SAX  is  similar  to  A  SPY  (?)  (315). 
V       A  SBC  -AX      A  SBC  AX 


But,  =  8*L*°  (?)  (388)  ;  and  —  =  —  (?)  (323, 3) 

A  SQR       SQ-SR^  PY      SP^ 


TT  V        SB  •  SC  SA       SA  •  SB  -  SC  ,  . 

Hence,  — ,  =  —  -  (Ax.  6), 

V       S     •  SR   SP       SP  -  S     •  SR  ^ 


Q.E.D, 


BOOK  VII 


321 


637.  THEOREIVL  In  any  polyhedion  the  number  of  edges  increased 
by  two  is  equal  to  the  number  of  vertices  increased  by  the  number  of 
faces. 


Given  :  A  polyhedron  ;  E  =  number  of  edges  ;  F=  number 
of  faces ;  F  =  number  of  vertices. 


To  Prove : 


=  V+F. 


Proof:  Suppose  the  surface  of  the  polyhedron  is  put  to- 
gether, face  by  face. 

For  one  face,  E—  V  (154).     (Begin  with  the  base.) 
By  attaching  an  adjoining  face,  the  number  of  edges  is  one 
greater  than  the  number  of  vertices. 

That  is,  for  two  faces,  E=  F  +  1. 

Similarly,  for  three  faces,  E  =  V  +  2, 

and,  for  four  faces,  E=  F+  3, 

and,  for  five  faces,  E  =  F  -{-  4, 


and,  for  n  faces,  E  =  V  +  (n  —  1), 

and,  for  F—  1  faces,  E=  F+  (F-  2). 

By  attaching  the  last  face,  neither  the  number  of  edges 
nor  the  number  of  vertices  is  increased. 

That  is,  for  F  faces,  #  =  F  +  F  -  2. 

.•.  for  the  complete  solid,  E+  2  =  F-f  F  (Ax.  2).      Q.E.D. 


322 


SOLID  GEOMETRY 


638.  THEOREM.   In  any  polyhedron  the  difference  between  the  num- 
ber of  edges  and  the  number  of  faces,  is  two  less  than  the  number  of 
vertices,  that  is,  E  -  F  =  V-  2.     (See  637.) 

639.  THEOREM.   The  sum  of  all  the  face  angles  of  any  polyhedron 
is  equal  to  4  right  angles  multiplied  by  two  less  than  the  number  of 
vertices,  that  is,  8  ^  =  ( F-  2)  4  rt.  A  =  ( V-  2)  360°. 


t      J 


Given  :  A  polyhedron ;  E  =  number  of  edges ;  F=  number 
of  faces ;  F  =  number  of  vertices. 

.  To  Prove :  Sum  of  all  the  face  A  =  (F  —  2)  4  rt.  A, 

ors4  =  (F-2)  360°. 

Proof:  Considering  the  faces  as  separate  polygons,  it  is 
obvious  that  each  edge  is  a  side  of  two  polygons,  that  is,  the 
number  of  sides  of  the  several  faces  =  2  E. 

.•.  the  number  of  vertices  of  all  the  polygons  =  2  E  (154). 

Suppose  an  exterior  Z  formed  at  each  of  these  2  E  vertices. 

Then  the  sum  of  the  exterior  A  of  each  face  =  4  rt.  A  (?). 

Sum  of  int.  and  ext.  A  at  each  vertex  =  2  rt.  A  (?). 

.-.  the  int.  A  4-  ext.  A  at  all  the  2  E  vertices  =  4  E  rt.  A. 

But,  the  ext.  A  of  all  the     F  faces       =  4  F  rt.  A  (?). 

Hence,  the  int.  A  of  these  polygons 

=  4^  rt.  A-4FTt.  ^(?). 

=  (#-*')•  4  rt.  A 
But  E-F=  F-2  (?)  (638). 

.-.  s4  =  (F-2)  4  rt.  z§  =  (F-2)  360°  (Ax.  6).        Q.E.D. 


BOOK  VII  323 

REGULAR  AND  SIMILAR  POLYHEDRONS 

640.  A  regular  polyhedron  is  a  polyhedron  whose  faces  are 
equal  regular  polygons  and   whose   polyhedral   angles   are 
all  equal. 

Similar  polyhedrons  are  polyhedrons  which  have  the  same 
number  of  faces  similar  each  to  each  and  similarly  placed, 
and  which  have  their  homologous  polyhedral  angles  equal. 

641.  THEOREM.    There  cannot  exist  more  than  five  kinds  of  regular 
polyhedrons. 

Proof :  The  faces  must  be  equilateral  A,  squares,  regular 
pentagons,  or  some  other  regular  polygons  (?)  (640). 

There  must  be  at  least  three  faces  at  each  vertex  (?)  (580). 

Sum  of  the  face  A  at  each  vertex  must  be  <  360°  (?)  (564). 
Let  us  now  discuss  the  possible  regular  polyhedrons. 

I.  Each  Z  of  an  equilateral  A  =  60°  (?).     Hence,  we  may 
form  a  polyhedral  Z  by  placing  3  equilateral  A  at  a  vertex, 
or  4  at  a  vertex,  or  5  at  a  vertex.     But  not  6  at  a  vertex  (?). 
That  is,  only  three  regular  polyhedrons  can  be  formed  having 
equilateral  triangles  for  fac*es. 

II.  Each  Z  of  a  square  =  90°  (?).     Hence,  we  may  form  a 
polyhedral  Z  by  placing  3  squares  at  a  vertex.     But  not  4 
at  a  vertex  (?).     That  is,  only  one  regular  polyhedron  can  be 
formed  having  squares  for  faces. 

III.  Each   Z  of   a   regular   pentagon  =  108°.  (?)    (164). 
Hence,  we  may  form  a  polyhedral  Z  by  placing  3  regular 
pentagons  at  a  vertex.     But  not  4  at  a  vertex  (?).     That  is, 
only  one  regular  polyhedron  can  be  formed  having  regular 
pentagons  for  faces. 

IV.  Each  Z  of  a  regular  hexagon  =  120°  (?).     Hence,  no 
polyhedral  Z  can  be  formed  by  hexagons  (?). 

Consequently,  there  can  be  no  more  than  five  kinds  of 
regular  polyhedrons, — three  kinds  bounded  by  triangles,  one 
kind  by  squares,  and  one  by  pentagons.  Q.E.D. 


324  SOLID   GEOMETRY 

642.    The  names  of  the  regular  polyhedrons. 


NAMES 

TOTAL 
NUMBER 

OF  FACES 

NUMBER  OF 
FACES  AT 
EACH  VERTEX 

•  KINDS  OF  FACES 

Regular  tetrahedron 

4 

3 

Equilateral  triangles 

Regular  hexahedron  (cube) 

6 

3 

Squares 

Regular  octahedron 

8 

4 

Equilateral  triangles 

Regular  dodecahedron 

12 

3 

Regular  pentagons 

Regular  icosahedron 

20 

5 

Equilateral  triangles 

A 


REGULAR 
TETRAHEDRON 


REGULAR  REGULAR  REGULAR 

OCTAHEDRON    DODECAHEDRON    ICOSAHEDRON 


BOOK  VII 


325 


DIRECTIONS.  —  The  regular  polyhedrons  may  be  constructed  as  fol- 
lows :  Mark  on  cardboard  figures  similar  to  the  drawings  on  page  324, 
but  much  larger;  with  a  knife  cut  the  dotted  lines  half  through  and  the 
solid  lines  entirely  through ;  fold  along  the  dotted  lines,  closing  the  solids 
up  and  forming  the  figures  as  illustrated  on  page  324;  paste  strips  of 
paper  along  the  edges  to  hold  them  in  position. 

643.  THEOREM.  In  two  similar 
polyhedrons : 

I.  Homologous  edges  are  propor- 
tional. 

II.  Homologous  faces  are  to  each 
other  as  the  squares  of  any  two  ho- 
mologous edges. 

III.  Total  surfaces  are  to  each 
other  as  the  squares  of  any  two  ho- 
mologous edges. 

Proof:  I.    Homologous  faces  are  similar  (?)  (640). 
AB         BC         CD         DH         AE         BF 


A'B'     B'C'     C'D' 

Face   DG        DH 


D'H' 
2 


II. 


III. 


A'E'     B'F' 

face  AH        AE2 


=  etc.  (?)  (823,  3). 


Face  D'G' 
Face  DG 


ITJ! 


j)'H'2     face  A'H 
face  AH      face  GE 


=  etc.  (?)  (390). 
feach  being 


Face  D'G' 
Total  surface  of  AG 


=  etc.  J 
lace  G'E'  equal  to 


face  DG        DH' 


Total  surface  of  A'o'      face  Df  G1 


TV* 

-^-=etc.  (301). 

A'E'2 


644.   THEOREM.    If  a  pyramid  is  cut  by  a  plane  parallel  to  the  base, 
the  pyramid  cut  away  is  similar  to  the  original  pyramid.    (Def .  of  640.) 


Ex.  1.     Show  that  the  theorem  of  637  holds  true  in  the  case  of  a  cube. 
Ex.  2.     Show  that  the  theorem  of  639  holds  true  in  the  case  of  a  cube. 
Ex.  3.     Show  that  the  theorems  of  637  and  639  are  true  in  the  case  of 
a  regular  octahedron. 


326 


SOLID   GEOMETRY 


O' 


645.   THEOREM.    Two  similar  tetrahedrons  are  to  each  other  as  the 
cubes  of  any  two  homologous  edges. 

Given:  Similar  tetrahedrons 
O-ABC  and  o'-A's'c'-,  whose 
volumes  =  F  and  F'. 

To  Prove: 
F  :  F'  =  AB*  =  J7F3=  etc. 

Proof :  Trihedral  Z  A  =  tri- 
hedral Z  A'  (?)  (640). 

V_         AB  •  AC 
"    F'~ 


AO 


A'B'.  A' C'- A'O1 
AB       AC       AO 


(636). 


But 


AC 

T^C'" 

v_ 

'"'   F' 


.!'£' 
^jg 

I7!*7 

^45 


^I'O' 


^10 
A'O'' 

AB 


AB 


A'B'    A'B'    A'B' 


(643,1). 
x.  6), 


That  is,       F  :  F'  = 


etc.       Q.E.D. 


Ex.  1.  If,  in  the  figure  of  645,  A  0  =  4  and  A'O'-  1,  what  is  the  ratio 
of  the  total  surfaces  of  the  tetrahedrons  ?  Of  their  volumes  ? 

Ex.  2.  Two  homologous  edges  of  two  similar  polyhedrons  are  2  and  3. 
What  is  the  ratio  of  their  total  surfaces? 

Ex.  3.  Two  homologous  edges  of  two  similar  tetrahedrons  are  2  and  5. 
The  area  of  the  total  surface  of  the  less  is  28,  and  its  volume  is  40.  Find 
the  area  of  the  total  surface  and  the  volume  of  the  other,  ^-ji^-  \i „ 

Ex.  4.  Show  that  the  theorems  of  637  and  639  are  true  in  the  cases  of 
regular  dodecahedrons  and  regular  icosahedrons. 

Ex.  5.  A  pyramid  whose  volume  is  F  and  altitude  is  h  is  bisected  by 
a  plane  parallel  to  the  base.  Find  the  distance  of  this  plane  from  the 
vertex. 

Ex  6.  Find  the  total  surface  of  a  regular  tetrahedron  whose  edges 
are  each  equal  to  4  in.  ^XO-\/3  -  l(o\/^  ^  V]%T- 

Ex.  7.  Find  the  total  surface  of  a  regular  octahedron  whose  edges 
are  each  equal  to  18  in.  f  I  \fe  X  f  •,  .  (^f  ff  \g  .=.  n 


BOOK   VII 


327 


646.  THEOREM.  Two  similar  polyhedrons  can  be  decomposed  into 
the  same  number  of  tetrahedrons  similar  each  to  each  and  similarly 
placed. 

Given:    (?). 

To  Prove:    (?). 

Proof:   Suppose  diagonals  I   I"  I    EF/I  B' 

drawn  in  every  face  of  AT, 
except  the  faces  containing 
vertex  A,  dividing  the  faces 
into  A.  (The  figure  shows 
only  SV.)  Suppose  lines 
drawn  from  A  to  the  several 
vertices  of  these  A.  (The  figure  shows  only  AS, 


Obviously,  this  process  divides  the  solid  (by  planes)  into 
tetrahedrons,  each  of  which  has  a  vertex  at  A. 

Then  construct  homologous  lines  in  solid  ArTr.  There 
will  evidently  be  as  many  lines  in  AfTf  as  in  AT  and  as  many 
tetrahedrons,  and  these  will  be  similarly  placed. 

Now,  in  the  tetrahedrons  A-SVR  and  A'-S'V'R^  A  AVR  is 
similar  to  A  A'v'lt'  i  A  ARS  is  similar  to  A  ArR's' ;  A  SVR 
is  similar  to  A  s'v'R1  (?)  (327). 

AV  VR          VS  RS          AS 


Also, 


,000 

(3  8' 


AV 


VS 


AS 


'  ArVl    "  V'S1       AfS 

Hence,  A  ASV  is  similar  to  A  Ars'vf  (?)  (318). 

Also,  the  trihedral  A  R  and  Rf  are  =  ;  8  and  Sr  are  =  ; 
F  and  v'  are  =  ,  etc.  (?)  (561). 

.*.  the  two  tetrahedrons  are  similar. 

Furthermore,  after  removing  these  tetrahedrons,  the  re- 
maining polyhedrons  are  similar  (Def.  640). 

By  the  same  process  other  tetrahedrons  may  be  removed 
and  proved  similar,  and  the  process  continued  until  the 
polyhedrons  are  completely  decomposed  into  tetrahedrons 
similar  each  to  each  and  similarly  placed.  Q.E.D. 


328 


SOLID   GEOMETRY 


647.   THEOREM.     The  volumes  of  two  similar  polyhedrons  are  to 
each  other  as  the  cubes  of  any  two  homologous  edges. 

Given:  Similar  polyhe- 
drons.^ and  A'T'I  volumes 
F  and  F';  AR  and  AfRf,  any 
two  homologous  edges. 

To  Prove:   (?) 

Proof:  These  solids  may 
be  decomposed,  etc.  (646). 
Denote  vol.  of  tetrahedrons 
of  AT  by  w,  x,  y,  z,  etc. ;  of 
A'T'  by  w',  «',  y',  zr,  etc. 


Bf 


W 


rp, 

Then,  — 


;  & 

'  V1 


A'R'* 


;  etc.  (?)  (645). 


Hence,  —,  =  -t -  =  £ •  =  etc.  (Ax.  1).      Therefore 

y*»i'  S**f  /til 


w' 


etc. 


" 


(gel).    That  is  -    = 
w'  +  xr  +  y1  +  etc.       w1  V1 


AR 


(Ax.  6). 

Q.E.D. 


648.   THEOREM.     The  volumes  of  two  similar  pyramids  are  to  each 
other  as  the  cubes  of  their  altitudes.     (Explain.) 

FORMULAS  OF  BOOK  VII 


Let  B  =  area  of  base. 

b  =  area  of  upper  base. 

E  =  number  of  edges. 
e,  e'  =  homologous  edges. 

F  —  number  of  faces. 

h  =  altitude. 

L  =  lateral  area. 
Parallelepiped     .     . 

Prism 

Regular  pyramid  . 
Pyramid  .... 
Frustum  of  pyramid 
Polyhedron  .  .  . 
Similar  polyhedrons  T  ••  T'  =  e'2  • 


P  =  perimeter  of  base. 
Pr  =  perimeter  of  right  section. 
p  =  perimeter  of  upper  base. 
s  =  slant  height. 
T  =  total  area. 
V  =  volume  or  number  of  vertices. 

V=B   h. 

£  =  !*,. -edge;    T  =  L  +  2  B  ;     V=B-h. 
Li=\P-8\    T=  L  +  B. 


;  Sum  of  face  A  =(F-  2)360° 
;    V-.V  =  e* :e's. 


BOOK  VII 


329 


ORIGINAL  EXERCISES 

1.  What  plane  through  the  vertex  of  a  given  tetrahedron  will  divide 
it  into  two  equivalent  parts?     Prove. 

2.  The  area  of  the  base  of  any  pyramid  is  less  than  the  sum  of  the 
lateral  faces. 

[Draw  the  altitudes  of  the  lateral  faces  and  the  projections  of  the  alti- 
tudes upon  the  base.] 

3.  Three  of  the  edges  of  a  parallelepiped  that  meet  in  a  point  are 
also  the  lateral  edges  of  a  pyramid.     What  part  of 

the  parallelepiped  is  this  pyramid? 

4.  A  plane  is  passed  containing  one  vertex  of 
a  parallelepiped  and  a  diagonal  of  a  face  not  con- 
taining that  vertex.      What  part  of    the   volume 
of  the  parallelepiped  is  the  pyramid  thus  cut  off? 

5.  Any  section  of  a  tetrahedron  made  by  a  plane 
parallel  to  two  opposite  edges,  is  a  parallelogram. 

Given  :  Section  DEFG  II  to  OA  and  EC. 
To  Prove :  DEFG  is  a  O . 

Proof:  EF  is  II  to  OA,  DG  is     to  OA  (?).     Also 
DE  is  II  to  BC  and  GF  is  ||  to  BC  (?)  etc. 

6.  The  three  lines  that  join  the  midpoints  of  the  opposite  edges  of  a 
tetrahedron  meet  in  a  point  and  bisect  one  another. 

Given  :  LM,  PQ,  RS,  the  three  lines,  etc. 
To  Prove  :  (?). 

Proof :  Join  PS,  SQ,  QR,  PR.     PS  is  II  to  and 
=  %BC;  RQ  is  I!  to  and  =  \  BC.  (Explain.)  A* 

Similarly,  discuss  LM  and  SR. 

7.  A  pyramid  having  one  of  the  faces  of  a  cube  for  its  base  and  the 
center  of  the  cube  for  its  vertex,  contains  one  sixth  of  the  volume  of 
the  cube.  O 

8.  A  plane  containing  an  edge  of  a  regular  tetra- 
hedron and  the  midpoint  of  the  opposite  edge, 

(a)  contains  the  medians  of  two  faces  ; 
(5)  is  perpendicular  to  the  opposite  edge ; 

(c)  is  perpendicular  to  these  two  faces; 

(d)  contains  two  altitudes  of  the  tetrahedron. 


330  SOLID   GEOMETRY 

9.    The   altitude  of  a  regular   tetrahedron    meets  the  base  at  the 
point  of  intersection  of  the  medians  of  the  base. 

10.  The  altitude  of  a  regular  tetrahedron  =  ^V6  times  the  edge. 

11.  The  altitudes  of  a  regular  tetrahedron  meet  at  a  point. 

12.  The  lines  joining  the  vertices  of  any  tetrahedron  to  the  point  of 
intersection  of  the  medians  of  the  opposite  face  meet  in  a  point  that 
divides  each  line  into  segments  in  the  ratio  3  : 1. 

Given:  OM,  CR,  two  such  lines. 
To  Prove  :  The  four  such  lines  meet,  etc. 
Proof  :  OM  and  CR  lie  in  the  plane  determined 
by  OC  and  point  D,  the  midpoint  of  AB. 

:.  OAT  and  CR  intersect.     Draw  RM.  &4 


'        "DM 

:.  RM  is  II  to  OC  (?). 

/.  A  DMR  and  DCO  are  similar  (?)  ;  and  DR:DO  =  RM:  OC  (?). 

Thus  RM  =  i  OC.     (Explain.) 

Also  &  PRM  and  OPC  are  similar  (?). 

Hence,  OP :  PM=  OC :  RM  =3:1.) 

and  CP  :  PR  =  OC :  RM  =  3  : 1.  | '  OxPlamO  Q.E.D. 

NOTE.     This  point  P  is  called  the  center  of  gravity  of  the  tetrahedron. 

13.  There   can    be  no  polyhedron    having  seven  V 
edges  and  only  seven. 

14.  The  planes  bisecting  the  dihedral   angles  of 
any  tetrahedron    meet  in   a  point  that  is  equally 
distant  from  the  faces. 

15.  The  lines  perpendicular  to  the  faces  of  any 

tetrahedron,  at  the  centers  of  the  circles  circumscribed  about  the  faces, 
meet  in  a  point  that  is  equally  distant  from  the  vertices. 

Proof  :  RX  and  SY  are  loci  of  points,  etc.  (526). 

Plane  MN,  JL  to  AB  at  M,  the  midpoint  of  AB, 
is  the  locus  of  points,  etc. 

.'.  RX  and  SY  lie  in  MN  and  intersect  at  0,  etc. 

16.  If    a    plane   be   passed  through    the   mid- 
points of  the  three  edges  of  a  parallelepiped  that 
meet  at  a  vertex,  what  part  of  the  whole  solid  is 
the  pyramid  thus  cut  off  ? 


BOOK   VII 


331 


17.  The  plane  bisecting  a  dihedral  angle  of  a  tetrahedron  divides  the 
opposite  edge  into  two  segments  proportional  to  the  areas  of  the  faces 
that  form  the  dihedral  angle. 

18.  Two  tetrahedrons  are  similar  if  a  dihedral  angle  of  one  equals  a 
dihedral  angle  of  the  other  and  the  faces  forming  these  dihedral  angles 
are  respectively  similar. 

19.  If  from  any  point  within  a  regular  tetrahedron  perpendiculars  to 
the  four  faces  are  drawn,  their  sum  is  constant  and 

equal  to  the  altitude  of  the  tetrahedron.  :£ 


20.  To  construct  a  regular  tetrahedron  upon  a 
given  edge. 

Construction  :  Upon  AB,  construct  an  equilateral 
A  ABC.  Erect  ED  JL  to  plane  of  A  ABC,  at  D> 
the  center  of  circumscribed  O.  Take  V  on  ED  such 
that  A  V=BV  =  CV=  AB,  etc. 

21.  To  construct   a  regular  hexahedron  upon  a 
given  edge. 

Construction :  Upon  A  B,  construct  a  square 
ABCD.  At  the  vertices  erect  Js  =  AB  and  join  the 
extremities,  etc. 

22.  To  construct  a  regular  octahedron  upon  a 
given  edge. 

Construction :  Upon  A  B,  construct  a  square 
ABCD.  At  M,  the  center  of  the  square,  erect  XX' 
JL  to  plane  of  ABCD.  On  XX'  take  MV.=MV  = 
MD.  Draw  the  edges  from  V  and  V . 

Statement :   VV  is  a  regular  octahedron. 

Proof:  The  right  A  DMV,  DMC,  DMV,  are 
equal.  (Explain.) 

Thus  the  12  edges  are  equal  and  the  8  faces  are 
equal.  (Explain.) 

Figures  AVCV,  DVBV,  ABCD  are  equal 
squares.  (Explain.) 

Then,  pyramids  V-ABCD,  D-A  FCF',etc.,  are  equal  and  the  6  poly- 
hedral angles  are  equal.     (Explain.)       .'.  Etc. 

23.  To  pass  a  plane  through  a  cube  so  that  the  section  will  be  a 
regular  hexagon. 


332 


SOLID   GEOMETRY 


24.    To  pass  planes  through  three  given  lines  in  space,  no  two  of 
which  are  parallel,  which  shall  inclose  a  parallelepiped. 


25.  Find  the  lateral  area  and  the  total  area  of  a  regular  pyramid 
whose  slant  height  is  20  in.  and  whose  base  is  a  square,  1  ft.  on  a  side. 

26.  Find  the  volume  of  a  pyramid  whose  altitude  is  18  in.  and  whose 
base  is  an  equilateral  triangle  each  side  of  which  is  8  in. 

27.  A  regular  hexagonal  pyramid  has  an  altitude  of  9  ft.  and  each 
edge  of  the  base  is  6  ft.     Find  the  volume.    ' 

28.  The  base  of  a  pyramid  is  an  isosceles  triangle  whose  sides  are  14, 
25,  25,  and  the  altitude  of  the  pyramid  is  12.     Find  its  volume. 

29.  The  altitude  of  the  frustum  of   a  pyramid  is  25,  and  the  bases 
are  squares  whose  sides  are  4  and  10,  respectively.      Find  the  volume 
of  the  frustum. 

30.  The  frustum  of  a  regular  pyramid  has  hexagons  for  bases  whose 
sides  are  5  and  9,  respectively.     The  slant  height  of  the  frustum  is  14. 
Find  its  lateral  area.     Find  its  total  area. 


31.  The  altitude  of  a  regular  pyramid  is  15,  and 
each  side  of   its  square  base  is  16.     Find  the  slant 
height,  the  lateral  edge,  the  total  area,  and  the  volume. 

OA2  =  OD2  +  DA2  =  (15)2  +  (8)2  =  289. 

/.  A O  =  17.       OC2  =  OA2  +  AC2  =289  +  64  =  353. 

:.  OC=  V353  =  18.78+.  ' 

32.  The  slant  height  of  a  regular  pyramid  is  39, 
the  altitude  is  36,  and  the  base  is  a  square.     Find  the 
lateral  area  and  volume. 

33.  The  lateral  edge  of  a  regular  pyramid  is  37 
and  each   side   of  the   hexagonal  base   is   12.     Find 
the  slant  height,   the  lateral  area,  total    area,  and 
volume. 

AD  =  6  V3. 


In  rt.  A  A  CD,  CD  =  12,  A  C  =  6, 

In  rt.  A  A  CO,  CO  =  37,  A  C  =  6,    .'.  AO-  VH£Ja. 

In  rt.  A  CDO,  CO  =  37,  CD  =  12, .'.  OD  =  35,  etc. 


BOOK  VII 


333 


34.  Find  the  total  area  and  volume  of  a  regular 
tetrahedron  whose  edge  is  six. 

The  four  faces  are  equal  equilateral  &.  .'.  AO  =  AC 
=  3V3;  /.^Z>  =  V3~and  CD  =  2V3. 

Hence,  OD  =  2  v'o.     Area  of  any  face  =  9  \/3,  etc. 

35.  Find  the  total  area  and  volume  of  a  regular  tetrahedron  whose 
edge  is  10. 

36.  Find  the  total  area  and  volume  of  a  regular  hexahedron  whose 

edge  is  8. 

37.  Find  the  total  area  and  volume  of  a  regular 
octahedron  whose  edge  is  16. 

The  8  faces  are  equal  equilateral  A.  A  0  =  8  V3.  In 
&ADO,  one  finds  OD  =  SV2.  The  volume  of  the 
octahedron  =  volumes  of  two  pyramids,  etc. 

38.  Find  the  total  area  and  volume  of  a  regular 
octahedron  whose  edge  is  18. 

39.  The   altitude   of  a  regular  pyramid  is  16  and  each  side  of  the 
square  base  is  24.     Find  the  lateral  area  and  volume. 

40.  The  slant  height  of  a  regular  pyramid  is  26  and  its  base  is  an 
equilateral  triangle  whose  side  is  20\/37    Find  the  total  area  and  volume. 

41.  The  altitude  of  a  regular  pyramid  is  29  and  its  base  is  a  regular 
hexagon  whose  side  is  10.     Find  the  total  area  and  volume. 

42.  Find  the  total  area  and  volume  of  a  regular  tetrahedron  whose 
edge  is  18. 

43.  Find  the  total  area  and  volume  of  a  regular  octahedron  whose 
edge  is  20. 

44.  If  the   edge  of  a  regular  tetrahedron  is  e,  show  that  the  total 
area  ise2v/3  and  the  volume  is  j1^  es  V2  . 

45.  If  the  edge  of  a  regular  octahedron  is  e,  show  that  the  total  area 
is  2  e2V3   and  the  volume  is  £  es  V2  . 

46.  A  pyramid  whose  base  is  a  square  9  in.  on  a  side,  contains  360  cu. 
in.     Find  its  height. 

47.  A  pyramid  has  for  its  base  a  hexagon  whose  side  is  7|  units  and 
the  pyramid  contains  675  cu.  units.     Find  the  altitude. 

48     The  volume  of  a  regular  tetrahedron  is  144  A/2 ;  find  its  edge. 
49.   The  volume  of  a  regular  octahedron  is  243  V^;  find  its  edge. 


334  SOLID  GEOMETRY 

50.  The  volume  of  a  square  pyramid  is  676  cu.  in.  and  the  altitude  is 
a  foot.    Find  the  side  of  the  base.     Find  the  lateral  area. 

51.  The  altitude  of  the  Great  Pyramid  is  480  ft.  and  its  base  is  764 
ft.  square.     It  is  said  to  have  cost  $  10  a  cu.  yd.  and  $  3  more  for  each 
sq.  yd.  of  lateral  surface  (considered  as  planes).     What  was  the  cost? 

52.  The  total  surface  of  a  regular  tetrahedron  is  324  V3  sq.  in. ;  find 
its  volume. 

53.  The  base  of  a  pyramid  is  a  rhombus  whose  diagonals  are  7  m. 
and  10  m.     Find  the  volume  if  the  altitude  is  15  m. 

54.  The  areas  of  the  bases  of  the  frustum  of  a  pyramid  are  3  sq.  m. 
and  27  sq.  m.     The  volume  is  104  cu.  m.     Find  the  altitude. 

55.  The   base  of  a  pyramid   is   an  isosceles  right    triangle    whose 
hypotenuse  is  8.     The  altitude  of  the  pyramid  is  15.     Find  the  volume. 

56.  The  altitude  of  a  square   pyramid,  each  side  of  whose  base  is 
6  ft.,  is  10  ft.     Parallel  to  the  base  and  2  ft.  from  the  vertex  a  plane  is 
passed.     Find  the   area  of  the  section.     Find  the  volumes  of  the  two 
pyramids  concerned,  and  hence  find  the  volume  of  the  frustum. 

57.  Find  the  area  of  the  section  of  a  triangular  pyramid  each  side  of 
whose  base  is  8  in.  and  whose  altitude  is  18  in.,  made  by  a  plane  parallel 
to  the  base  and  a  foot  from  the  vertex. 

58.  The  altitude  of  a  frustum  of  a  pyramid  is  6,  and  the  areas  of  the 
bases  are  20  sq.  in.  and  45   sq.  in.     Find  the  altitude  of  the  complete 
pyramid.     Find  the  volume  of  this  frustum  by  two  distinct  methods. 

59.  A  granite  monument  in  the  form  of  a  frustum  of  a  pyramid, 
having  rectangular  bases  one  of  which  is  8  ft.  wide  and  12  ft  long,  and 
the  other  6   ft.   wide,   is  30  ft.  high.     It  is  surmounted  by  a  granite 
pyramid   having  the  same  base  as  the  less  base  of  the  frustum,  and 
10  ft.  in   height.     Find   the  entire   volume.     If  one   cu.   ft.   of  water 
weighs  62|  Ib.  and  granite  is  three  times  as  heavy  as  water,  what  is  the 
weight  of  the  entire  monument? 

60.  If  a  square  pyramid  contains  40  cu.  in.  and  its  altitude  is  15  in., 
find  the  side  of  its  base. 

61.  A  church  spire  in  the  form  of  a  regular  hexagonal  pyramid  whose 
base  edge  is  8  ft.  and  whose  altitude  is  75  ft.  is  to  be  painted  at  the  rate 
of  18^  per  square  yard.     Find  the  cost. 

62.  Find  the  edge  of  a  cube  whose  volume  is  equal  to  the  volumes  of 
two  cubes  whose  edges  are  4  and  6. 


BOOK  vil  335 

i 

63.  The  base  of  a  certain  pyramid   is  an  isosceles  trapezoid  whose 
parallel  sides  are  20  ft.  and  30  ft.  and  the  equal  sides  are  each  13  ft. 
Find  the  volume  of  the  pyramid  if  its  altitude  is  12  yards. 

64.  The  lateral  edge  of  the  frustum  of  a  regular  square  pyramid  is 
53  and  the  sides  of  the  bases  are  10  and  66.     Find  the  altitude,  the  slant 
height,  the  lateral  area,  and  the  volume. 

65.  The  sides  of  the  base  of  a  triangular  pyramid  are  33,  34,  65,  and 
the  altitude  of  the  pyramid  is  80.     Find  its  volume. 

66.  The  sides  of  the  base  of  a  tetrahedron  are  17,  25,  26,  and  its 
altitude  is  90.     Find  its  volume. 

67.  If  there   are  l^-  cu.  ft.  in  a  bushel,  what   is   the   capacity  (in 
bushels)  of  a  hopper  in  the  shape  of  an  inverted  pyramid,  12  ft.  deep 
and  8  ft.  square  at  the  top  ? 

68.  In  the  corner  of  a  cellar  is  a  pyramidal  heap  of  coal.     The  base 
of  the  heap  is  an  isosceles  right  triangle  whose  hypotenuse  is  20  ft.  and 
the  altitude  of  the  heap  is  7  ft.     If  there  are  35  cu.  ft.  in  a  ton  of  coal, 
how  many  tons  are  there  in  this  heap  ? 

69.  How  many  cubic  yards  of  earth  must  be  removed  in  digging  an 
artificial  lake  15  ft.  deep,  whose    base  is  a  rectangle  180  x  20  ft.  and 
whose  top  is  a  rectangle  216  x  24  ft.  ?     [The  frustum  of  a  pyramid.] 


70.  One  pair  of  homologous  edges  of  two  similar  tetrahedrons  are 
3  ft.  and  5  ft.     Find  the  ratio  of  their  surfaces.     Of  their  volumes. 

71.  A  pair  of  homologous  edges  of  two  similar  polyhedrons  are  5  in. 
and  7  in.     Find  the  ratio  of  their  surfaces.     Of  their  volumes. 

72.  The  edge  of  a  cube  is  3.     What  is  the  edge  of  a  cube  twice  as 
large  ?     Four  times  as  large  ?    Half  as  large  ? 

73.  An  edge  of  a  tetrahedron  is  6.     What  is  the  edge  of  a  similar 
tetrahedron  three  times  as  large?     Eight  times  as  large?    Nine  times  as 
large  ?    One  third  as  large  ? 

74.  An  edge  of  a  regular  icosahedron  is  3  in.     What  is  the  edge  of 
a  similar  solid  five  times  as  large  ?     Ten  times  as  large  ?     Fifty  times  as 
large?    A  thousand  times  as  large? 

75.  The  edges  of  a  trunk  are  2  ft.,  3  ft.,  5  ft.     Another  trunk  is  twice 
as  long  (the  other  edges  2x3  ft.).     How  do  their  volumes  compare? 
A  third  trunk  has  each  dimension  double  those  of  the  first.     How  does 
its  volume  compare  with  the  first  ?     How  do  their  surfaces  compare  ? 


336  SOLID   GEOMETRY 

76.  If  the  altitude  of  a  certain  regular  pyramid  is  doubled,  but  the 
base  remains  unchanged,  how  is  the  volume  affected?     If  each   edge 
of  the  base  is  doubled,  but  the  altitude  unchanged,  how  is  the  volume 
affected  ?    If  the  altitude  and  each  edge  of  the  base  are  all  doubled,  how 
is  the  volume  affected? 

77.  If  the  slant  height  (only)  of  a  regular  pyramid  is  doubled,  how 
is  the  lateral  area  affected  ?     If  each  edge  of  the  base  is  doubled,  how  is 
the  lateral  area  affected?    If  both  are  doubled,  what  is  the  effect? 

78.  A  pyramid  is  cut  by  a  plane  parallel  to  the  base  and  bisecting  the 
altitude.     What  part  of  the  entire  pyramid  is  the  less  pyramid  cut  away 
by  this  plane  ? 

79.  The  volume  of  a  certain  pyramid,  one  of  whose  edges  is  7,  is  686. 
Find  the  volume  of  a  similar  pyramid  whose  homologous  edge  is  8. 

80.  A  certain  polyhedron  whose  shortest  edge  is  2  in.  weighs  40  Ib. 
What  is  the  weight  of  a  similar  polyhedron  whose  shortest  edge  is  5  in.  ? 

81.  An  edge  of  a  polyhedron  is  5  in.  and  the  homologous  edge  of  a 
similar  polyhedron  is  7  in.     The  entire  surface  of  the  first  is  250  sq.  in. 
and  its  volume  is  375  cu.  in.     Find  the  entire  surface  and  volume  of  the 
second. 

82.  Find  the  edge  of  a  cube  whose  volume  equals  that  of  a  rectangular 
parallelepiped  whose  edges  are  3  x  4  x  18. 

83.  A  pyramid  and  an  equivalent  prism  have  equivalent  bases.     How 
do  their  altitudes  compare  ? 

84.  A  pyramid  and  a  prism  have  the  same  altitude  and  equivalent 
bases.     Compare  their  volumes. 

85.  Solve:  V=%B.hiorB.     Fork. 

86.  Solve  :  L  =  \  P  .  s  for  P.    For  s. 

87.  Solve  :  L  =  \  (P  +p) .  s  for  s. 

88.  A  prism  whose  altitude  is  8  and  whose  base  is  an  equilateral  tri- 
angle whose  side  is  9  in.  is  transformed  into  a  regular  pyramid  whose 
base  is  10  in.  square.     Find  its  altitude. 

89.  An  altitude  of  a  pyramid  is  10  m.     How  far  from  the  vertex  will 
a  plane  parallel  to  the  base  divide  the  pyramid  into  two  equivalent  parts  ? 

90.  The  altitude  of   a  pyramid    is  12,  and   two  planes  are  passed 
parallel  to  the  base  and  dividing  the  pyramid  into  three  equivalent  parts. 
At  what  distances  from  the  vertex  are  they  ? 


BOOK   VIII 

CYLINDERS,    CONES 
CYLINDERS 

649.  A  cylindrical  surface  is  a  surface  generated  by  a 
moving  straight  line  which  continually  intersects  a  given 
curved  line  in  a  plane,  and  which  is  always  parallel  to 
a  given  straight  line  not  in  the  plane  of  the  curve. 

The  generating  line  is  the  generatrix.  The  directing 
curve  is  the  directrix. 

An  element  of  a  cylindrical  surface  is  the  generating  line 
in  any  position. 


/  ULn 


CYLINDRICAL  RIGHT  OBLIQUE  CYLINDER 

SURFACE  CIRCULAR  CYLINDER  OF  REVOLUTION 

CYLINDER 

650.  A  cylinder  is  a  solid  bounded  by  a  cylindrical  surface 
and  two  parallel  planes. 

The  bases  of  a  cylinder  are  the  parallel  plane  sections. 

The  lateral  area  of  a  cylinder  is  the  area  of  the  cylindrical 
surface  included  between  the  planes  of  the  bases. 

The  total  area  of  a  cylinder  is  the  sum  of  the  lateral  area 
and  the  areas  of  the  bases. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance 
between  the  planes  of  the  bases. 

337 


338  SOLID   GEOMETRY 

651.  A  right   cylinder  is  a  cylinder  whose  elements  are 
perpendicular  to  the  planes  of  the  bases. 

A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 

An  oblique  cylinder  is  a  cylinder  whose  elements  are  not 
perpendicular  to  the  planes  of  the  bases. 

A  right  circular  cylinder  is  a  right  cylinder  whose  base  is 
a  circle. 

A  cylinder  of  revolution  is  a  cylinder  generated  by  the 
revolution  of  a  rectangle  about  one  of  its  sides  as  an  axis. 

Similar  cylinders  of  revolution  are  cylinders  generated  by 
similar  rectangles  revolving  on  homologous  sides. 

652.  A  right  section  of  a  cylinder  is  a  section  made  by 
a  plane  perpendicular  to  all  the  elements. 

A  plane  is  tangent  to  a  cylinder  if  it  contains  one  element 
of  the  cylindrical  surface  and  only  one,  however  far  it  may 
be  extended. 

A  prism  is  inscribed  in  a  cylinder  if  its  lateral  edges  are 
elements  of  the  cylinder  arid  the  bases  of  the  prism  are  in- 
scribed in  the  bases  of  the  cylinder. 

A  prism  is  circumscribed  about  a  cylinder  if  its  lateral 
faces  are  tangent  to  the  cylinder  and  the  bases  of  the  prism 
are  circumscribed  about  the  bases  of  the  cylinder. 

PRELIMINARY   THEOREMS 

653.  THEOREM.     Any  two  elements  of  a  cylinder  are  equal  and 
parallel.     (See  524  and  511.) 

654.  THEOREM.     A  line  drawn  through  any  point  in  a  cylindrical 
surface,  parallel  to  an  element,  is  itself  an  element.     (See  92.) 

655.  THEOREM.     A  right  circular  cylinder  is  a  cylinder  of  revolu- 
tion. 

Ex.  If  a  plane  be  defined  as  a  surface  generated  by  a  moving  straight 
line,  what  would  the  directrix  be? 


BOOK  VIII 


339 


THEOREMS   AND   DEMONSTRATIONS 

656.  THEOREM.     Every  section  of  a  cylinder  made  by  a  plane  con 
taining  an  element  is  a  parallelogram. 

Given  :  Cylinder  AE  ;  plane 
CE  containing  element  CD. 

To  Prove  :  CE  is  a  O. 

Proof :  At  E  draw  EF  II  to 
CD  in  plane  CE.  Also,  EFis  an 
element  of  the  cylinder  (654). 

.*.  EF  is  the  intersection  of 
the  plane  and  the  cylindrical 
surface  (?)  (482). 

Also,  CFis  II  to  DE  (?)  (500). 

/.  CDEFis  aO  (?)  (126). 

657.  THEOREM.     The  bases  of  a  cylinder  are  equal. 
Given:  (?).     To  Prove :  (?) 

Proof :  Suppose  R,  8,  and  T 
any  three  points  in  the  perim- 
eter of  base  AC.  Draw  ele- 
ments RR',  ss',  TT'.  Also, 
draw  RS,  8T,  RT,  R'S',  S'T', 
R'T'.  Now, 


Q.E.D. 


8S'is  =  and  II  to  TT 

.'.  RS'  is  a  O,  RT'  is  a  O,  ST'  is  a  O  (?)  (135). 

.'.BS=R'S';  8T=  S'T'',RT=R'T'(1W)',  A  RST=  A  fl's'r'  (?). 

.*.  base  u4C  may  be  placed  upon  base  BD  so  that  ft,  /s,  and  T 
coincide  with  R1,  -s',  and  Tf,  respectively.  But  S  is  any  point 
on  the  perimeter ;  hence,  every  point  on  perimeter  of  AC  will 
coincide  with  a  corresponding  point  on  perimeter  of  BD. 

Therefore,  base  AC  =  base  BD  (?)  (28).        Q.E.D. 


340 


SOLID  GEOMETRY 


658.  COR.     Parallel  plane  sections  of  a  cylinder  (cutting  all  the 
elements)  are  equal. 

659.  COR.     Any  section  of  a  circular  cylinder  made  by  a  plane 
parallel  to  the  base  is  a  circle. 

660.  THEOREM.     Every  section  of  a  right  cylinder  made  by  a  plane 
containing  an  element  is  a  rectangle  (?). 

661.  THEOREM.     If  a  regular  prism  be  inscribed  in,  or   circum- 
scribed about,  a  right  circular  cylinder  and  the   number   of   sides 
of  the  base  be  indefinitely  in- 
creased, the  lateral  area  of  the 

cylinder  is  the  limit  of  the  lat- 
eral area  of  the  prism. 

Given :  A  regular  prism 
inscribed  in  and  a  regular 
prism  circumscribed  about 
a  right  circular  cylinder ; 
the  lateral  area  of  the  cyl- 
inder =  L,  and  of  the  prisms, 
Lt  and  ic,  respectively. 

To  Prove:  That  as  the 
number  of  sides  of  the  bases 
of  the  prisms  is  indefinitely 
increased,  L  is  the  limit  of  both  iz  and  Lc. 

Proof :  If  the  number  of  sides  of  the  bases  of  the  prisms 
is  indefinitely  increased,  their  perimeters  will  approach 
the  circumference  of  the  base  of  the  cylinder  as  a  limit  (?). 

Hence,  it  is  obvious  that  the  lateral  area  of  the  cylinder 
is  the  limit  of  the  lateral  area  of  either  prism.  Q.E.D. 


Ex.  1.  If  the  cylindrical  surface  of  a  cylinder  be  cut  along  an  element, 
and  this  surface  be  placed  in  coincidence  with  a  plane,  what  plane 
geometrical  figure  will  it  become  ? 

Ex.   2.   What  two  lines  determine  the  size  of  aright  circular  cylinder? 


BOOK  VIII 


341 


662.  THEOREM.     If  a  prism  having  a  regular  polygon  for  a  base 
be  inscribed  in,  or  circumscribed  about,  any  circular  cylinder  and  the 
number  of  the  sides  of  the  base  of  the  prism  be  indefinitely  in- 
creased, the  volume  of  the  cylinder  is  the  limit  of  the  volume  of 
the  prism. 

Proof :  If  the  number 
of  sides  of  the  base  of 
either  prism  be  indefi- 
nitely increased,  the  area 
of  the  base  of  the  prism 
will  approach  the  area  of 
the  base  of  the  cylinder 
(440,  II). 

.'.it  is  obvious  that  the 
volume  of  the  cylinder  is 
the  limit  of  the  volume  of 
either  prism.  Q.E.D. 

663.  THEOREM.     The  lateral  area  of  a  right  circular  cylinder  is 
equal  to  the  product  of  the  circumference  of  the  base  by  an  element. 

Given  :  A  right  circular  cylinder, 
the  circumference  of  whose  base  = 
(7,  and  whose  element  =  E. 

To  Prove :    Lateral  area  L  =  C  •  E. 

Proof :  Inscribe  in  the  cylinder  a 
regular  prism,  the  perimeter  of 
whose  base  is  P,  whose  lateral  edge 
is  E,  and  whose  lateral  area  is  L'. 

Then  L'  =  P  •  E  (?).  If  the  num- 
ber of  sides  of  the  base  of  the  prism 
is  indefinitely  increased,  L1  will 
approach  L  as  a  limit  (?).  P  will  approach  C  as  a  limit  (?). 

P  •  E  will  approach  C  •  E  as  a  limit. 

,'.L=C-E  (?)  (242).  Q.E.D. 


342 


SOLID  GEOMETRY 


NOTE.  The  lateral  area  of  an  oblique  circular  cylinder  is 
equal  to  the  product  of  the  perimeter  of  a  right  section  of 
the  cylinder  by  an  element. 

The  right  section  of  an  oblique  circular  cylinder  is  not  a 
circle.  The  right  section  of  an  inscribed  prism,  having  a 
regular  polygon  for  a  base,  is  not  a  regular  polygon.  Since 
elementary  geometry  does  not  deal  with  curves  other  than 
the  circle,  any  proof  or  application  of  this  theorem  is  omitted. 


664. 


V664.\    THEOREM.     The  volume  of  a  circular  cylinder  is  equal  to  the 
prdcttict  of  its  base  by  its  altitude. 

Given:  A  circular  cylinder 
whose  base  =  .B,  altitude  =  A,  and 
volume  =  F. 

To  Prove:   V=B  -  h. 

Proof  :  Inscribe  a  prism  having 
a  regular  polygon  for  its  base, 
whose  base  =  Br  and  volume  =  V1  . 

Then,  F'=  £'  •  A(?)  (603).     If 
the  number  of  sides  of  the  base 
of  the  prism   is  indefinitely  increased,  V1  approaches  F  as 
a  limit  (?),  Br  approaches  B  as  a  limit  (?),  and 
B'  •  h  approaches  B  -  h  as  a  limit. 

/.  V  =  B  •  h  (?)  (242).  Q.E.D. 

FORMULAS 

Let  B  —  area  of  base.  h  =  altitude.  T  =  total  area. 

E  =  element.  L  =  lateral  area.  F  =  volume. 

C  =  circumference  of  base.  R  =  radius  of  base. 

665.     Lateral  area  of  right  circular  cylinder,  L  =  C  •  E  (?). 


666.  Total  area  of  right  circular  cylinder,  T=i  +  2  B  (?). 
/.  T=27rM  +  27TE2  (?).      .-.  T=  21TJB(A+.B). 

667.  Volume  of  circular  cylinder,  V—B  •  h  (?). 


BOOK   VIII 


343 


668.     THEOREM.     Of  two  similar  cylinders  of  revolution  : 

~i.   The  lateral  areas  are  to  each  other  as  the  squares  of 
altitudes  or  as  the  squares  of  the  radii  of  their  bases. 

II.   The  total  areas  are  to  each  other  as  the  squares  of 
altitudes  or  as  the  squares  of  the  radii  of  their  bases. 

III.  The  volumes  are  to  each 
other  as  the  cubes  of  their  alti- 
tudes or  as  the  cubes  of  the 
radii  of  their  bases. 

Given  :  Two  similar  cylin- 
ders of  revolution  whose  lat- 
eral areas  —  L  and  l\  whose 
total  areas  =  T  and  t\ 
whose  volumes  =  V  and  v  ; 


their 
their 


whose  altitudes  =  H  and  A,  and  whose  radii  are  E  and  r. 

To  Prove  : 

I.    L:  l  =  H2:  A2=,R2:  r2. 

II.    T:  t  =  H*:  A2=E2-  r2. 
III.    Vi  v  =  H*:  A3  =  jR3.  ,* 

Proof:    The   generating   rectangles  of  the   cylinders 
similar  (?)  (651). 

' 


are 


Hence,  H  +  R  :  h+  r=  H  :  h  =  R  :  r  (?)  (301). 


L         2 


EH       R    H       H  H        II 


I       27rrh 


-.-  =  -.-  =±        ?_.  (Explain.) 

^^         } 


III       -  =  7rK2ff 


R(g+ff)  _  R     H+R  =  H    H 
r(A-f-r)   ~     r     h  +  r        h    h 

7T2  7?2 


r2     h 


r3 


Q.E.D. 


344  SOLID   GEOMETRY 

ORIGINAL   EXERCISES   (NUMERICAL) 

TT  =  3f  1  bu.  =  2150.42  cu.  in.  1  gal.  =  231  cu.  in. 

In  a  cylinder  of  revolution, 

1.  If  R  =  5  in.,  h  =  14  in.,  find  L;  T\   V. 

2.  If  12  =  7m.,  A  =  10m.,  find  Z;  T7;    F. 

3.  If  12  =  4f  ft.,  A  =  18  ft.,  find  Z;  T\  V. 

4.  If  '12  =  6  in.,  L  =  792  sq.  in.,  find  h ;  77;  F. 

5.  If  R  =  4,  !T  =  352,  find  A;  Z;  I7. 

6.  If  12  =  2,  F=  22,  find  A;  L;  T. 

7.  If  A  =  5.6,  Z  =  352,  find  12;  7";  F. 

8.  If   A  =  9,  !F  =  440,  find  12;  L;  F. 

9.  If  h  =  9£,  F=  66,  find  12;  Z;  T7. 

10.  If  Z  =  440,  T7  =  1672,  find  12 ;  h ;  F. 

11.  If  L  =  198,  F  =  594,  find  12;  A;  T7. 

12.  How  many  square  inches  of  tin  will  be  required  to  make  a  cylin- 
drical pail  10  in.  in  diameter   and   a  foot  in  height,  without   any  lid? 
How  many  gallons  will  it  contain? 

13.  The  diameter  of  a  well  is  5|  ft.  and  the  water  is  14  ft.  deep.    How 
many  gallons  of  water  in  the  well? 

14.  In  a  cylinder  of  revolution  generated  by  a  rectangle  30  x  14  in. 
revolving  about  its  shorter  side  as  an  axis,  find  L ;  T ;  F. 

15.  In  a  cylinder  of  revolution  generated  by  the  rectangle  of  No.  14, 
revolving  about  its  longer  side  as  an  axis,  find  L  ;  T ;  F. 

16.  A  Cylindrical  vessel  9  in.  high,  closed  at  one  end,  required  361f 
sq.  in.  of  tin  in  its  construction.     Find  its  radius. 

17.  A  cylindrical  pail  12  in.  high  holds  exactly  two  gallons.     Find  12. 

18.  How  many  cu.  ft.  of  metal  in  a  hollow  cylindrical  tube  42  ft.  long, 
whose  outer  and  inner  diameters  are  10  in.  and  6  in.,  respectively? 

19.  A  tunnel  whose  cross  section  is  a  semicircle  18  ft.  high  is  one 
mi.  long.     How  many  cu.  yd.  of  material  were  removed  in  the  excavation  ? 

20.  An  irregular  stone  is  placed  in  a  cylindrical  vessel  a  in.  in  diameter 
and  partly  full  of  water.     The  water  rises  b  in.     Find  volume  of  stone. 

21.  A  rod  of  copper  18  ft.  long  and  2  in.  square  at  the  end  is  melted 
and  formed  into  a  wire  £  in.  in  diameter.     Find  the  length  of  the  wire. 

22.  If  a  cylindrical  bushel  measure  has  for  its  altitude  the  diameter 
of  the  base,  find  the  altitude. 


BOOK  VIII  345 

CONES 

l  A  conical  surface  is  a  surface  generated  by  a  moving 
straight  line  that  continually  intersects  a  given  curve  in  a 
plane,  and  passes  through  a  fixed  point  not  in  this  plane. 

The  generating  line  is. the  generatrix.  The  directing  curve 
is  the  directrix.  The  fixed  point  is  the  vertex  of  the  conical 
surface. 

An  element  of  a  conical  surface  is  the  generating  line  in 
any  position. 


ob*-l  m         I        m*        3^- 


CONICAL       RIGHT        OBLIQUE       CONE  OF         FRUSTUM 
SURFACE    CIRCULAR       CONE      REVOLUTION     OF  A  CONE 
CONE 

670.  A  cone  is  a  solid  bounded  by  a  conical  surface  and  a 
plane  cutting  all  the  elements. 

The  base  of  a  cone  is  its  plane  surface. 

The  lateral  area  of  a  cone  is  the  area  of  the  conical  surface. 

The  total  area  of  a  cone  is  the  sum  of  the  lateral  area  and 
the  area  of  the  base. 

The  altitude  of  a  cone  is  the  perpendicular  distance  from 
the  vertex  to  the  plane  of  the  base. 

671.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 

The  axis  of  a  circular  cone  is  the  line  drawn  from  the 
vertex  to  the  center  of  the  base. 

A  right  circular  cone  is  a  circular  cone  whose  axis  is  per- 
pendicular to  the  plane  of  the  base. 


346  SOLID   GEOMETRY 

An  oblique  circular  cone  is  one  whose  axis  is  oblique  to  the 
plane  of  the  base. 

A  cone  of  revolution  is  a  cone  generated  by  the  revolution 
of  a  right  triangle  about  one  of  the  legs  as  an  axis. 

Similar  cones  of  revolution  are  cones  generated  by  the 
revolution  of  similar  right  triangles  revolving  about  homol- 
ogous sides. 

The  slant  height  of  a  cone  of  revolution  is  any  one  of  its 
elements. 

672.  A  frustum  of  a  cone  is  the  portion  of  a  cone  between 
the  base  and  a  plane  parallel  to  the  base. 

The  altitude  of  a  frustum  of  a  cone  is  the  perpendicular 
distance  between  the  planes  of  its  bases.  The  slant  height 
of  a  frustum  of  a  cone  is  the  portion  of  an  element  included 
between  the  bases.  The  lateral  area  of  a  frustum  is  the  area 
of  its  curved  surface.  The  total  area  of  a  frustum  is  the 
sum  of  the  lateral  area  and  the  area  of  the  bases. 

673.  A  plane  is  tangent  to  a  cone  if  it  contains  one  ele- 
ment of  the  conical  surface  and  only  one,  however  far  it  may 
be  extended. 

A  pyramid  is  inscribed  in  a  cone  if  its  base  is  inscribed 
in  the  base  of  the  cone,  and  its  vertex  is  the  vertex  of 
the  cone. 

A  pyramid  is  circumscribed  about  a  cone  if  its  base  is  cir- 
cumscribed about  the  base  of  the  cone,  and  its  vertex  is  the 
vertex  of  the  cone. 

The  frustum  of  a  pyramid  is  inscribed  in,  or  circumscribed 
about,  the  frustum  of  a  cone  if  the  bases  of  the  pyramid 
are  inscribed  in,  or  circumscribed  about,  the  bases  of  the 
cone. 

The  midsection  of  a  frustum  of  a  cone  is  the  section  made 
by  a  plane  parallel  to  the  bases,  and  midway  between  them. 


BOOK  VIII  347 


PRELIMINARY   THEOREMS 


674.  THEOREM.     The  elements  of   a  right   circular  cone  are  all 
equal.     (See  520,  II.) 

675.  THEOREM.     A  right  circular  cone  is  a  cone  ot  revolution. 

The  altitude,  any  element,  and  the  radius  of  the  base  of  a 
right  circular  cone  form  a  right  triangle,  and  in  the  same 
cone  all  such  triangles  are  equal  (53). 

676.  THEOREM.     The  altitude  of  a  cone  of  revolution  is  the  axis  of 
the  cone. 

677.  THEOREM.     A  straight  line  drawn  from  the  vertex  of  a  cone 
to  any  point  in  the  perimeter  of  the  base  is  an  element.     (See  39.) 

678.  THEOREM.     The  lateral   edges  of  a  pyramid   inscribed  in  a 
cone  are  elements  of  the  cone.     (See  677.) 

679.  THEOREM.     The  lateral  faces   of  a  pyramid   circumscribed 
about  a  cone  are  tangent  to  the  conical  surface.     (Explain.) 

680.  THEOREM.    The  slant  height  of  a  regular  pyramid  circum- 
scribed about  a  right  circular  cone  is  the  same  as  the  slant  height  of  the 
cone. 

681.  THEOREM.   The  slant  height  of  the  frustum  of  a  regular  pyra- 
mid circumscribed  about  the  frustum  of  a  right  circular  cone  is  the 
same  as  the  slant  height  of  the  frustum  of  the  cone.     (See  680.) 

682.  THEOREM.    The  radius  of  the  mid-section  of  a  frustum  of  a 
right  circular  cone  is  equal  to  half  the  sum  of  the  radii  of  the  bases. 

Proof :  The  radius  of  the  mid-section  is  the  median  of  a 
trapezoid  whose  bases  are  the  radii  of  the  bases  of  the 
frustum.  That  is,  m  =  £  (B  +  r).  (See  149.) 


Ex.  1.  What  two  lines  determine  the  size  of  aright  circular  cone? 
What  two  lines  determine  the  total  area? 

Ex.  2.  Find  the  slant  height  of  a  right  circular  cone  whose  altitude 
is  8  and  whose  radius  is  6. 


V 


348 


SOLID   GEOMETRY 


THEOREMS  AND  DEMONSTRATIONS 

683.  THEOREM.   Any  section  of  a  cone  made  by  a  plane  passing 
through  the  vertex  is  a  triangle. 

Given:  Cone  O-AB;  plane  OCD. 

To  Prove  :   Section  OCD  is  a  A. 

Proof:  Draw  straight  lines  OC, 
OD,  in  plane  OCD.  They  are  ele- 
ments (?)  (677). 

/.  OC  and  OD  compose  the  inter- 
section of  the  plane  and  the  conical 
surface  (?)  (482). 

Also  CD  is  a  straight  line  (?). 

.'.  OCD  is  a  A  (?)  (23).       Q.E.D. 

684.  THEOREM.    Any  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  is  a  circle. 

Given:  Cone  o  — AB;  circle  C 
its  base;  section  A'B'  11  to  base. 

To  Prove :  A'B'  also  a  O. 

Proof:  Draw  the  axis  OC  and 
pass  planes  OCD,  OCE  intersecting 
the  base  in  CD,  CE  respectively, 
and  the  section  in  C'D',  CfEr. 

In  A  OCD  and  OCE,  D'C'  is  II 
to  DC;  C'E'  is  II  to  CE  (?)  (500). 

/.  A  OC'Dr  is  similar  to  A  OCD; 
A  OC'E'  is  similar  to  A  OCE  (316). 


ocf 

OC 


C'D' 
CD 


^-  =  11^- (323,  3).  /.^-  = 

OC        CE    v  CD         CE 


But  CD=  CE  (?)  (200).     .-.  C'D'  =  C'E'  (Ax.  3).    That  is, 
all  points  on  the  perimeter  of  A'B'  are  equally  distant  from  C!. 
.'.  perimeter  of  section  A'B'  is  a  circumference  (?)  (192). 
Hence,  A'B'  is  a  O  (?)  (193).  Q.E.D. 


BOOK  VIII  349 

685.   THEOREM.     If  a  regular  pyramid  be  inscribed  in,  or  circum 
scribed  about,  a  right  circular  cone  and  the  number  of  sides  of  the  base 
be  indefinitely  increased,  the  lateral  area  of  the  cone  is  the  limit  of  the 
lateral  area  of  the  pyramid.     (See  Fig.  A.) 

Demonstration  is  similar  to  that  of  661. 


FIG.  A  FIG.  B 

686.  THEOREM.     If  a  pyramid  having  a  regular  polygon  for  a  base 
be  inscribed  in,  or  circumscribed  about,  any  circular  cone  and  the  number 
of  sides  of  its  base  be  indefinitely  increased,  the  volume  of  the  cone  is 
the  limit  of  the  volume  of  the  pyramid.    (See  Fig.  B.) 

Demonstration  is  similar  to  that  of  662. 

687.  THEOREM.     If  a  frustum  of  a  regular  pyramid  be  inscribed 
in,  or  circumscribed  about,  the  frustum  of  a  right  circular  cone  and  the 
number  of  sides  of  the  bases  be  indefinitely  increased,  the  lateral  area 
of  the  frustum  of  the  cone  is  the  limit  of  the  lateral  area  of  the 
frustum  of  the  pyramid. 

688.  THEOREM.     If  the  frustum  of   a  pyramid   having   regular 
polygons  for  its  bases  be  inscribed  in,  or  circumscribed  about,  a  frustum 
of  any  circular  cone  and  the  number  of   sides  of  the  bases  of  the 
frustum  be  indefinitely  increased,  the  volume  of  the  frustum  of  the 
cone  is  the  limit  of  the  volume  of  the  frustum  of  the  pyramid. 


350 


SOLID   GEOMETRY 


689.  THEOREM,     The  lateral  area  of  a  right  circular  cone  is  equal 
to  half  the  product  of  the  circumference  of  the  base  by  the  slant 
height. 

Given  :  Right  circular  cone 
O-AD,  the  circumference  of 
whose  base  =  C  and  whose 
slant  height  =  s. 

To  Prove : 

Lateral  area  =  J  C  •  s. 
Proof :  Circumscribe  a  regu- 
lar pyramid  and   denote   the 
lateral   area   by   Lf    and    the 
perimeter  of   the   base  by  P. 
Slant  height  OA  =s  (?)  (680). 
Then  £' =  J  P  •  «  (?)  (617). 

Now  indefinitely  increase  the  number  of  sides  of  the  base 
of  the  pyramid  and, 

Lr  will  approach  L  as  a  limit  (?)  (685). 
P  will  approach  C  as  a  limit  (?)  (440,  I). 
£  P-s  will  approach  J  C*s  as  a  limit  (?). 
Hence,  L  =  J  c  •  s  (?)  (242).  Q.E.D. 

690.  THEOREM.     The  lateral  area  of  the  frustum  of  a  right  circu- 
lar cone  is  equal  to  half  the  sum  of  the  circumferences  of  the  bases 
multiplied  by  the  slant  height. 

Given:  Frustum  of  right 
circular  cone,  whose  lateral 
area  is  L\  whose  slant  height 
is  s\  and  the  circumferences 
of  whose  bases  are  C  and  c. 

To  Prove:  L  =  J  (c  +  c)  •  s. 

Proof :  Circumscribe  a 
frustum  of  a  regular  pyramid 
and  denote  its  lateral  area  by 
L',  the  perimeters  of  its  bases  by  P  and  p. 


BOOK  VIII 


351 


The  slant  height  of  frustum  of  pyramid  =  s  (?)  (681). 

Now,  i'  =  J(P  +;>)•«(?)  (618). 

Indefinitely  increase  the  number  of  the  sides  of  the  bases 
of  the  frustum  of  the  pyramid,  and 

Lf  will  approacli  L  as  a  limit  (?), 
also  P  will  approach  c  as  a  limit  (?), 
and  p  will  approach  c  as  a  limit  (?), 
and  J  (P  -\-p)  -s  will  approach  J  (c  +  <?)  •  s  as  a  limit  (?). 
Hence,  £  =  |  (c  +  c)  •  *  (?).  Q.E.D. 

691.   THEOREM.     The  volume  of  a  circular  cone  is  equal  to  one 
third  the  product  of  the  area  of  the  base  by  the  altitude. 

Given:  Circular  cone  O-AD, 
whose  volume  =  F;  area  of 
whose  base  =  B  ;  altitude  = 
OE=h. 

To  Prove:   F=  J  B  •  h. 

Proof  :  Circumscribe  (or 
inscribe)  a  pyramid  having  a 
regular  polygon  for  its  base. 
Denote  the  volume  of  the 
pyramid  by  F',  its  base  by  Bf, 
its  altitude  =  OE=h  (507). 


Indefinitely  increase  the  number  of  the  sides,  etc. 
Then  vr  will  approach  F  as  a  limit  (?). 
Also  Bf  will  approach  B  as  a  limit  (?)  (440,  II). 
%  Br  •  h  will  approach  J  B  •  Ti  as  a  limit  (?). 

Hence,  F=  J  B  -  h  (?).  Q.E.D. 


Ex.  If  the  conical  surface  of  a  cone  of  revolution  be  cut  along  an 
.element  and  the  conical  surface  be  placed  in  coincidence  with  a  plane, 
what  geometrical  figure  will  the  surface  be  ? 


352 


SOLID   GEOMETRY 


692.  THEOREM.  The  volume  of  the  frustum  of  a  circular  cone  is 
equal  to  one  third  the  product  of  the  altitude  by  the  sum  of  the  lower 
base,  the  upper  base,  and  a  mean  proportional  between  the  bases  of 
the  frustum. 

Given:  A  frustum  of  any 
circular  cone,  whose  volume 
=  F,  whose  bases  are  1*  and  &, 
whose  altitude  is  h. 

To  Prove: 

F=^  A[B  +  6  +  Vj3-6]. 

Proof  :  Inscribe  (or  circum- 
scribe) a  frustum  of  a  pyra- 
mid having  regular  polygons 
for  bases.  Denote  its  volume 
by  F',  bases  by  B1  and  5',  and  altitude  by  h. 

Then,  F'  =  J  h[Br  +  b'  +  VFTp]  (?)  (631). 

Indefinitely,  etc.       Vr  will  approach  F  as  a  limit  (?). 

Br  and  V  will  approach  B  and  b  as  limits  (?). 
V.B'  •  br  will  approach  Vl?  •  b  as  a  limit. 

J  h[Br  +  br  -f-  VV  •  b']   will  approach   |   A[JB  -f-  £  +  VlTT>]. 

Hence,         F  =  J-  A[>  +  fl  +  ViTT]  (?)  (242).  Q.E.D. 


Ex.  1.  Could  you  prove  the  theorem  of  690  by  inscribing  a  frustum 
of  a  pyramid  ?  Could  you  prove  the  theorem  of  692  by  circumscribing  a 
frustum  of  a  pyramid  ?  Give  reasons  for  your  answer. 

Ex.  2.  Could  you  prove  the  theorem  of  689  by  inscribing  a  pyramid? 
Could  you  prove  the  theorem  of  691  by  inscribing  a  pyramid  V  Give 
reasons. 

Ex.  3.  The  frustum  of  a  circular  cone  is  15  in.  high  and  the  bases 
are  circles  whose  radii  are  3  in.  and  5  in.  Find  the  volume. 

Ex.  4.  The  frustum  of  a  right  circular  cone  has  a  slant  height  of 
9  ft.  and  the  radii  are  5  ft.  and  7  ft.  Find  the  lateral  area  and  the  total 
area.  What  is  the  length  of  the  altitude  of  this  frustum? 


BOOK  VIII  353 

FORMULAS 

Let  B  =  area  of  base.  m  =  radius  of  mid-section  of  frustum. 

b  =  area  of  less  base.  R  =  radius  of  base. 

C  —  circumference  of  base.  r  —  radius  of  less  base. 

c  =  circumf.  of  less  base.  s  =  slant  height. 

h  =  altitude.  T  =  total  area. 

L  =  lateral  area.  V  =  volume. 

693.  Lateral  area  of  right  circular  cone, 


694.   Total  area  of  right  circular  cone,  T  =  TTKS  +  TrR2  (?). 

+  JB). 


695.   Volume  of  circular  cone,  V=  J  B  •  h  =  ^  TrR2  •  h  (?). 


696.   Lateral  area  of  frustum  of  right  circular  cone, 


/.  L  =ir(R  +  r)s.     Also,  L  =  7r(2  m)s  =  2  irms  (682). 

697.  Total  area  of  frustum  of  right  circular  cone, 
T=  TT(B  +  r)9 

.-.  T  =  ir[(JB 


698.   Volume  of  frustum  of  circular  cone, 


+  Trr2  + 


Ex.  1.  State  in  words  each  of  the  final  formulas  of  the  paragraphs 
of  this  page. 

Ex.  2.  Find  the  radius  of  a  circle  having  the  same  area  as  the  lateral 
area  of  a  cone  of  revolution  whose  radius  is  4  and  slant  height,  9. 


354 


SOLID  GEOMETRY 


1 699.  THEOREM.     Of  two  similar  cones  of  revolution : 

I.  The  lateral  areas  are  to  each  other  as  the  squares  of  their  alti- 
tudes, or  as  the  squares  of  their  radii,  or  as  the  squares  of  their  slant 
heights. 

II.  The  total  areas  are  to  each  other  as  the  squares  of  their  alti- 
tudes, or  as  the  squares  of  their  radii,  or  as  the  squares  of  their  slant 
heights. 

III.  The  volumes  are  to  each  other  as  the  cubes  of  their  altitudes, 
or  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their  slant  heights. 

Given:  Two  similar  cones 
of  revolution,  whose  respec- 
tive lateral  areas  are  L  and 
Z,  total  areas  are  T  and  £, 
volumes  are  V  and  v,  alti- 
tudes are  H  and  A,  radii  are 
R  and  r,  slant  heights  are 
S  and  s. 


To  Prove 


y      _L/ a   K  o 


TT        L  —          —  —  —  _ 


III. 

Proof:  The  generating 
Hence, 


L 

_H2_&_ 

s2 

l  = 

h2      r2 

s2' 

T 

H2      R2_ 

s2 

7" 

h2      r2 

«»' 

V 

HB     B3 

s3 

V 

hs      r3 

s3 

o  T»p  cirnilfiT 

(?\       •    H  —R  —  8  (9} 

.  dl  \5  oliUlitll 

^'}    "  h      r      s  U 

===      (?)  (301  and  Ax.  1). 
r+s       r      s      h 


L^TTRS^^R     S^H    H==n_=R_=S^         (Explain.) 
I       TTTS       r     s      h     h       h2      r2      s2 


II.   - 


r+s 


h~  h2      r2      s2 


(Explain.) 


±7rr2h       r2     h       h2     h 


\\ 


BOOK  VIII  355 

ORIGINAL  EXERCISES 


In  a  cone  of  revolution, 
<L   If   A  = 
2.   If   A  = 
3.   If£  = 
•^4.    If   h  = 
Xs.   If  A  = 
-  6.   If  £  = 

12, 
15, 

18, 
6, 
20, 

7, 

5=13, 

£  =  8, 
s  =  30, 

5=10, 

£=21, 

s=25, 

find 
find 
find 
find 
find 
find 

£. 
s. 
A. 
£;  Z;  T;  F. 
s;  Z;  5T;   F. 
A;  Z;  T;    F. 

7. 

If 

L  = 

4070, 

5  =  37, 

find 

£; 

A;  T\ 

F. 

8. 

If 

L  = 

46.64, 

R  =  2.8, 

find 

s; 

A;  r; 

F. 

9. 

If 

Z  = 

400, 

T  =  500, 

find 

s; 

A;  £; 

F. 

10. 

If 

T  = 

80  7T, 

£  =  5, 

find 

«; 

A;  Z; 

F. 

11. 

If 

T  = 

10  7T, 

s  =  3, 

find 

£; 

A;  Z; 

F. 

12. 

If 

F  = 

462, 

£  =  21, 

find 

A; 

s;  Z; 

T". 

13.    If  F=8Tf7,     A  =  3,      find£;  s;  i;  7\ 

-   14.   What  would  be  the  cost  at  10  j*  a  square  foot  of  painting  a  conical 
church  steeple,  112  ft.  high  and  30  ft.  in  diameter  at  the  base? 

15.  The  sides  of  an  equilateral   triangle  are  each  12  in.     Find  the 
lateral  surface,  total  surface,  and  volume  of  the  solid  generated  by  re- 
volving this  triangle  about  an  altitude  as  an  axis. 

16.  An  isosceles  right  triangle  whose  legs  are  each  8  is  revolved  about 
the  hypotenuse  as  an  axis.     Find  the  total  surface  and  volume  of  the 
solid  generated. 

17.  The  sides  of  an  equilateral  triangle  are  each  10.     Find  the  total 
surface  and  the  volume  of  the  solid  generated  by  revolving  this  triangle 
about  one  of  its  sides  as  an  axis. 

18.  Find  the  volume  of  a  cone  of  revolution  whose  slant  height  is 
16  and  lateral  area  is  192  TT. 

19.  Find  the  lateral  area  of  a  cone  of  revolution  whose  altitude  is  20 
and  volume  is  240  TT. 

20.  How  many  bushels  in  a  conical  heap  of  grain  whose  base  is  a 
circle  35  ft.  in  diameter,  and  whose  height  is  25  ft.  ? 

21.  A  regular   hexagon  whose  side  is  6  revolves  about   one  of  the 
longer  diagonals.     Find  the  surface  and  the  volume  of  solid  generated. 

22.  Find  the  volumes  of   the  right  circular  cones  inscribed  in  and 
circumscribed  about  a  regular  tetrahedron  whose  edge  is  a. 


356  SOLID   GEOMETRY 

23.  A  right  triangle  whose  legs  are  15  and  20  is  revolved  about  the 
hypotenuse  as  an  axis.  Find  the  surface  and  the  volume  of  the  solid 
generated.  

In  the  frustum  of  a  right  circular  cone, 
—  24.   If  h  =  8,      R  =  10,    r  =  4,  find  s;  Z;  T;  V. 

25.  If  h  =  30,      s  =  34,    r  =  5,  find  R;  L;  T;  V. 

26.  If  s  =  19|,  £  =  10J,  r  =  3,  find  h]  L-,  T;  F. 

27.  How  many  square  feet  of  tin  are  required  to  make  a  funnel  2  ft. 
long,  if  the  diameters  of  the  ends  are  20  and  56  in.,  respectively? 

28.  A  chimney  150  ft.  high  has  a  cylindrical  flue  3  ft.  in  diameter. 
The  bases  of  the  chimney  are  circles  whose  diameters  are  28  ft.  and  7  ft. 
Find  the  number  of  cubic  yards  of  masonry  in  the  chimney. 

29.  A  plane  is  passed  parallel  to  the  base  of  a  right  circular  cone 
and  f  the  distance  from  the  vertex  to  the  base.     Find  the  ratio  of  the 
smaller  cone  thus  formed  to  the  original  cone.     Compare  the  volume  of 
the  less  cone  with  the  frustum  formed. 

Original  cone  _  53  ^  _    , 

Less  cone      ~~  2^  ^  ~ 

Hence,  Original  cone  -  Less  cone  =  125-8    ?)  etc< 

Less  cone  8 

30.  The  altitude  of  a  cone  of  revolution  is  12  in.     What  is  the  alti- 
tude of  the  frustum  of  this  cone  that  shall  contain  one  fourth  the  volume 
of  the  whole  cone? 

31.  The  altitudes  of  two  similar  cylinders  of  revolution  are  3  and  5. 
What  is  the  ratio  of  their  lateral  areas  ?    Of  their  total  areas  ?    Of  their 
volumes  ? 

32.  The  altitudes  of  two  similar  cylinders  of  revolution  are  5  and  6, 
and  the  lateral  area  of  the  first  is  200.     Find  the  lateral  area  of  the 
second.     If  the  volume  of  the  first  is  500,  what  is  the  volume  of  the 
second  ? 

33.  The  total  areas  of  two  similar  cones  of  revolution  are  24  TT  and 
216  TT  and  the  radius  of  the  first  is  3.     Find  the  radius  of  the  second. 
The  slant  height  of  the  first  is  5.     Find  the  lateral  area  of  the  second. 
Find  the  altitude  of  the  first  and  the  volume  of  the  second. 

34.  The  volumes  of  two  similar  cones  of  revolution  are  27  TT  and 
343  TT.     The  altitude  of  the  first  is  9.     Find  the  altitude  of  the  second. 
Find  the  radius  of  the  base  of  each. 


BOOK  VIII  357 

35.  A  cone  of  revolution  whose  radius  is  10  and  altitude  20,  has  the 
same  volume  as  a  cylinder  of  revolution  whose  radius  is  15.     Find  the 
altitude  of  the  cylinder. 

36.  A  cylinder  of  revolution  whose  radius  is  8  and  altitude  30,  is 
formed  into  a  cone  of  revolution  whose  altitude  is  40.     Find  the  radius 
of  its  base. 

37.  The  heights  of  two  equivalent  cylinders  of  revolution  are  in  the 
ratio  of  4 : 9.     If  the  diameter  of  the  first  is  12  ft.,  what  is  the  diameter 
of  the  second  ? 

38.  A  cylinder  of  revolution  8  ft.  in  diameter  is  equivalent  to  a  cone 
of  revolution  7  ft.  in  diameter.     If  the  height  of  the  cone  is  16  ft.,  what 
is  the  height  of  the  cylinder? 


39.  Two  circular  cylinders  having  equal  altitudes  are  to  each  other 
as  their  bases. 

40.  Two  circular  cylinders  having  equal  bases  are  to  each  other  as 
their  altitudes. 

41.  Two  circular  cylinders  having  equal  bases  and  equal  altitudes  are 
equivalent. 

42.  Two  circular  cones  having  equal  altitudes  are  to  each  other  as 
their  bases. 

43.  Two  circular  cones  having  equal  bases  are  to  each  other  as  their 
altitudes. 

44.  Two  circular  cones  having  equal  bases  and  equal  altitudes  are 
equivalent. 

45.  If  the  altitude  of  a  right  circular  cylinder  is  equal  to  the  radius 
of  the  base,  the  lateral  area  is  half  the  total  area. 

46.  If  the  altitude  of  a  right  circular  cylinder  is  half  the  radius  of  the 
base,  the  lateral  area  is  equal  to  the  area  of  the  base. 

47.    If  the  slant  height  of  a  right  circular  cone  is  equal  to  the  diame- 
ter of  the  base,  the  lateral  area  is  double  the  area  of  the  base. 

48.  The  lateral  area  of  a  cone  of  revolution  is  equal  to  the  area  of 
a  circle  whose  radius  is  a  mean  proportional  between  the  slant  height  and 
the  radius  of  the  base. 

49.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to  the  area 
of  a  circle  whose  radius  is  a  mean  proportional  between  the  altitude  of 
the  cylinder  and  the  diameter  of  its  base. 


358  SOLID   GEOMETRY 

50.  What  relation  does  the  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  have  to  the  base?     Prove. 

51.  At  what  distance  from  the  vertex  of  a  right  circular  cone  whose 
altitude  is  h  must  a  plane  parallel  to  the  base  be  passed,  so  as  to  bisect 
the  lateral  area?     At  what  distance  must  it  be  passed  so  as  to  bisect  the 
volume? 

52.  What  does  the  volume  V  of  a  right  circular  cone  become,  if  the 
altitude  is  doubled  and  the  base  undisturbed  ?     Prove.     What  does  the 
volume  V  become  if  the  radius  of  the  base  is  doubled  but  the  altitude 
undisturbed?     Prove.     If  both  are  doubled ?     Prove. 

53.  The  intersection  of  two  planes  tangent  to  a  cylinder  is  a  line 
parallel  to  an  element. 

54.  The  intersection  of  two  planes  tangent  to  a  cone  is  a  line  through 
the  vertex. 

55.  One  straight  line  can  be  drawn  upon  a  cylindrical  surface  through 
a  given  point,  and  only  one. 

56.  If  two  cylinders  of  revolution  have  equivalent  lateral  areas,  their 
volumes  are  to  each  other  as  their  radii. 

57.  If  a  rectangle  be  revolved  about  its  unequal  sides  as  axes,  the 
volumes  of  the  two  solids  generated  are  inversely  proportional  to  the 
axes,  and  directly  proportional  to  the  radii  of  the  bases. 

58.  Show  that  the  formula  for  the  volume  of  a  circular  cone  can  be 
derived  from  the  formula  for  the  volume  of  a  frustum  of  a  circular  cone 
if  one  base  of  the  frustum  becomes  a  point. 

59.  Reduce  the  formula  for  the  volume  of  a  frustum  of  a  circular  cone 
if  the  radius  of  one  base  is  double  the  radius  of  the  other. 


60.  To  pass  a  plane  tangent  to  a  circular  cylinder  and  containing  a 
given  element. 

Construction :  Draw  a  line  in  plane  of  base  tangent  to  the  base  at  the 
end  of  the  given  element,  etc. 

61.  To  pass  a  plane  tangent  to  a  circular  cone  and  containing  a  given 
element. 

62.  To  pass  a  plane  tangent  to  a  circular  cylinder  and  through  a 
given  point  without  it. 

Construction :  From  the  point  draw  a  line  ||  to  an  element,  meeting 
the  plane  of  the  base.  From  this  point  of  intersection  draw  a  line  tan- 
gent to  the  base  of  the  cylinder.  Through  the  point  of  contact  draw  an 
element;  etc. 


BOOK  VIII  359 

63.  To  pass  a  plane  tangent  to  a  circular  cone  and  through  a  given 
point  without  it. 

Construction:   Connect  this  point  with  the  vertex  of  the  cone  and 
prolong  this  line  to  meet  the  plane  of  the  base,  etc. 

64.  To  divide  the  lateral  surface  of  a  cone  of  revolution  into  two 
equivalent  parts  by  a  plane  parallel  to  the  base. 

65.  Find  the  locus  of  points  at  a  given  distance  from  a  given  straight 
line. 

66.  Find  the  locus  of  points  equally  distant  from  two  given  points 
and  at  a  given  distance  from  a  straight  line.     Discuss. 

67.  Find  the  locus  of  points  at  a  given  distance  from  a  given  plane 
and  at  a  given  distance  from  a  given  line.     Discuss. 

68.  Find  the  locus  of  points  at  a  given  distance  from  a  given  cylin- 
drical surface  whose  generatrix  is  a  circle,  and  whose  elements  are  per- 
pendicular to  the  plane  of  the  circle. 

69.  Find  the  locus  of  all  lines  making  angles  equal  to  a  given  angle, 
with  a  given  line,  at  a  given  point. 

70.  Find  the  locus  of  all  lines  making  angles  equal  to  a  given  angle, 
with  a  given  plane,  at  a  given  point. 

71.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given  line 
and  equally  distant  from  two  given  planes.     Discuss. 

72.  Find  a  point,  X,  at  a  given  distance  from  a  given  line  and  equally 
distant  from  three  given  points.     Discuss. 


BOOK   IX 


THE    SPHERE 

700.  A  sphere  is  a  solid  bounded  by  a  surface,  all  points 
of  which  are  equally  distant  from  a  point  within,  called  the 
center.     The  surface  of  a  sphere  is  a  spherical  surface. 

A  radius  of  a  sphere  is  a  straight  line 
drawn  from  the  center  to  any  point  of  the 
surface. 

A  diameter  of  a  sphere  is  a  straight  line 
that  contains  the  center  and  has  its  ex- 
tremities in  the  surface. 

701.  THEOREM.     Every  plane  section  of  a  sphere  is  a  circle. 
Given :  Sphere  whose  center  is  O ; 

plane  MN  intersecting  sphere  in  AB. 

To  Prove:  The  figure  AB  is  a  O. 

Proof  :  Draw  OD  J_  to  plane  MN, 
meeting  the  plane  at  D.  Take  P  and 
Q,  any  two  points  on  the  perimeter  of 
the  section,  and  draw  DP,  DQ,  OP,  OQ. 

A  ODP  and  ODQ  are  rt.  A  (?). 

In  A  ODP  and  ODQ,  OD  =  OD  and  OP  =  OQ  (?)  (700). 

.-.the  A  are  equal  (?). 

Hence,  DP  =  DQ  (?). 

That  is,  all  points  of  the  perimeter  of  AB  are  equally  dis- 
tant from  D. 

.'.the  perimeter  of  AB  is  a  circumference  (?)  (192). 

Hence,  the  section  AB  is  a  circle  (?)  (193).        Q.E.D, 
360 


N 


BOOK  IX  361 

702.  A  great  circle  of  a  sphere  is  a  section  of  the  sphere 
made  by  a  plane  containing  the  center  of  the  sphere. 

A  small  circle  of  a  sphere  is  a  section  of  the  sphere  made 
by  a  plane  that  does  not  contain  the  center  of  the  sphere. 

The  axis  of  a  circle  of  a  sphere  is  the  diameter  of  the 
sphere  perpendicular  to  the  plane  of  the  circle. 

The  poles  of  a  circle  of  a  sphere  are  the  ends  of  its  axis. 

A  quadrant  (in  Spherical  Geometry)  is  one  fourth  of  the 
circumference  of  a  great  circle. 

Equal  spheres  are  spheres  having  equal  radii. 

703  A  plane  is  tangent  to  a  sphere  if  it  touches  the 
sphere  in  only  one  point.  Two  spheres  are  tangent  to  each 
other  if  they  are  tangent  to  the  same  plane  at  the  same 
point.  They  may  be  tangent  internally  or  externally. 

A  line  is  tangent  to  a  sphere  if  it  touches  the  sphere  in 
only  one  point  and  does  not  intersect  it. 

A  line  is  tangent  to  the  circle  of  a  sphere  if  it  lies  in  the 
plane  of  the  circle  and  touches  the  circle  at  only  one  point. 
In  all  cases  this  common  point  is  the  point  of  contact  or  point 
of  tangency. 

704.  A  sphere  is  inscribed  in  a  polyhedron  if  all  the  faces 
of  the  polyhedron  are  tangent  to  the  sphere. 

A  sphere  is  circumscribed  about  a  polyhedron  if  all  the 
vertices  of  the  polyhedron  are  in  the  spherical  surface. 

705.  The  distance  between  two  points  on  the  surface  of  a 
sphere  is  the  less  arc  of  a  great  circle  passing  through  them. 

The  distance  between  a  point  on  a  circle  of  a  sphere  and 
the  nearer  pole  of  the  circle  is  the  polar  distance  of  the  point. 

706.  The   angle   between  two   intersecting   curves  is   the 
angle  formed  by  their  tangents  at  the  point  of  intersection. 

A  spherical  angle  is  the  angle  between  the  circumferences 
of  two  great  circles  of  a  sphere. 


362  SOLID   GEOMETRY 

PRELIMINARY   THEOREMS 

707.  THEOREM.     All  radii  of  a  sphere  are  equal.     (See  700.) 

708.  THEOREM.     All  radii  of  equal  spheres  are  equal.     (See  702.) 

709.  THEOREM.     All  diameters  of  the  same  sphere  or  of   equal 
spheres  are  equal.     (See  Ax.  3.) 

710.  THEOREM.     All  great  circles  of  the  same  sphere  or  of  equal 
spheres  are  equal.     (See  196.) 

711.  THEOREM.     In  the  same  sphere  or  in  equal  spheres : 

I.    Equal  plane  sections  are  equally  distant  from  the  center. 
II.   Plane  sections  equally  distant  from  the  center  are  equal. 

III.  Of  two  unequal  plane  sections,  the  greater  is  at  the  less 
distance  from  the  center. 

IV.  Of  two  plane  sections  unequally  distant  from  the  center, 
the  section  at  the  greater  distance  is  the  less. 

In  each  case  the  diameters  of  the  sections  are  chords  of  great 
circles.     These  theorems  follow  from  221,  222,  223,  224. 

712.  THEOREM.     Two  great  circles  of  a  sphere  bisect  each  other. 
Because  they  have  a  common  diameter.     (See  204.) 

713.  THEOREM.     Every  great  circle  of  a  sphere  bisects  the  sphere 
and  the  spherical  surface.     (Proof  is  like  the  proof  of  204.) 

714.  THEOREM.     A  sphere  may  be  generated  by  the  revolution 
of  a  semicircle  about  the  diameter  as  an  axis. 

715.  THEOREM.     One  and   only  one   great   circle  can  be   drawn 
through  two  points,  not  the  ends  of  a  diameter,  on  the  surface  of  a 
sphere.     (See  493.) 

716.  THEOREM.     One  and  only  one  circle  can  be  drawn  through 
three  points  on  the  surface  of  a  sphere.     (See  493.) 


BOOK  IX  363 

717.  THEOREM.     A  point  is  without  a  sphere  if  its  distance  from 
the  center  is  greater  than  the  radius,  and  if  a  point  is  without  a 
sphere   its   distance  from  the  center  is    greater    than  the  radius. 

(See  700.) 

718.  THEOREM.     The  axis  of  a  circle  of  a  sphere  passes  through 
its  center.     (This  is  proved  in  the  proof  of  701.) 

THEOREMS   AND   DEMONSTRATIONS 

719.  THEOREM.     A  plane  perpendicular  to  a  radius  of  a  sphere  at 
its  extremity  is  tangent  to  the  sphere. 

Given :    Radius  OA  of  sphere  O,  and 
plane  MN  _L  to  OA  at  A. 

To  Prove  :  MN  tangent  to  the  sphere. 

Proof:    Take    any  point   X  in   MN 
(except  A)  and  draw  OX. 

OX  >  OA  (?)  (520,  I). 
.*.  X  lies  without  the  sphere  (717). 

Hence,  every  point  of  plane  MN,  except  A,  is  without  the 
sphere;  that  is,  plane  MN  is  tangent  to  the  sphere  (?)  (703). 

Q.E.D. 

720.  THEOREM.     A  plane  tangent  to  a  sphere  is  perpendicular  to 
the  radius  drawn  to  the  point  of  contact. 

Given :  Plane  MN  tangent  to  sphere  O  at  A ;  radius  OA. 
To  Prove  :  OA  is  _L  to  plane  MN. 

Proof:  Every  point  in  MN,  except  A,  is  without  the  sphere 
(?)  (703). 

Take  any  point  X  in  MN  and  draw  OX. 

Now,  OX  is  >  OA  (?)  (717).     That  is,  OA  is  the  shortest 
line  from  O  to  MN. 

.'.  OA  is  J_  to  MN  (?)  (520,  I).  Q.E.D. 


364 


SOLID   GEOMETRY 


721.   THEOREM.     A  line  tangent  to  a  sphere  lies  in  the  plane  tan- 
gent to  the  sphere  at  the  same  point. 

Proof:    The  line  is  _L  to  the   radius 
drawn  to  the  point  of  contact  (?)  (216). 
The  plane  also  is  -L  to  the  radius  (?). 
/.  the  line  is  in  the  plane  (502). 


722.  THEOREM.  At  a  point  on  the  surface 
of  a  sphere  there  can  be  only  one  tangent 
plane.  (Explain.) 


•N 


723.  THEOREM.  All  points  in  the  circumference  of  a  circle  of  a 
sphere  are  equally  distant  from  either  pole  ;  that  is,  the  polar  distances 
of  all  points  in  the  circumference  of  a  circle  are  equal. 

Given :  P  and  z,,  the  poles  of  O  C  on 
sphere  O,  and  great  (D  PAL,  PEL. 

To  Prove :    Arc  PA  =  arc  PB ; 

arc  AL  =  arc  BL. 

Proof :  Draw  the  axis  PL  meeting  plane 
of  O  C  at  C. 

Draw  AC,  AP,  BC,  BP. 

PC  is  _L  to  plane  DAB  (?)  (Def .  of  axis, 
702). 


.*.  chord  PA  =  chord  PB  (?). 
Hence,  arc  PA  =  arc  PB  (?). 
Likewise,  arc  AL  =  arc  BL. 


Q.E.D. 


724.  COR.     The  polar  distance  of  a  great  circle  is  a  quadrant. 

725.  THEOREM.     If  a  point  on  the  surface  of  a  sphere  is  at  the 
distance  of  a  quadrant  from  two  other  points  on  the  surface,  not  the 
ends  of  a  diameter,  it  is  the  pole  of  the  great  circle  containing  these 
two  points. 

Given :  P,  a  point,  and  R  and  2V,  two  other  points,  —  all  on 
the  surface  of  sphere  O  ;  arcs  PR  and  P7V,  quadrants ;  great 
circle  ARNB. 


BOOK  IX 


To  Prove :   P  is  the  pole  of  O  AENB. 

Proof :   Draw  the  radii  OP,  OB,  O-ZV. 
A  PON  and  FOR  are  rt.  A  (245). 
/.  PO  is  _L  to  plane  AB  (?)  (501). 
.-.  PO  is  the  axis  of  O  AENB  (702). 
.-.  P  is  the  pole  of  OAENB  (702).  Q.E.D, 


365 


72$.  THEOREM.  A  spherical  angle  is  measured  by  the  arc  of  a 
great  circle  having  the  vertex  of  the  angle  as  a  pole  and  intercepted  by 
the  sides  of  the  angle. 

Given:  Spherical  /.  A  VB;   arc  AB  of 
great  O  whose  pole  is  F,  on  sphere  O. 

To   Prove :  /.  A  VB    is   measured    by 
arc  AB. 

Proof  :   Draw  radii  OA,   OB,  OF,  and 

at  F  draw  VC  tangent  to  O  VA,  and  FD 
tangent  to  O  VB.  VB  is  a  quadrant  (724) . 
OF  is  _L  to  FD  (?),  and 
OF  is  _L  to  OB  (?)  (245). 
Likewise,  OF  is  J_  to  VC  and  to  OA  (?). 
.-.  FD  is  II  to  OB,  and  VC  is  II  to  OA  (?)  (93). 
.\^  CVD  =  Z.AOB  (?)  (515). 

And  Z.  CVD  is  the  spherical  Z.  AVB  (706). 
But  /.AOB  is  measured  by  arc  AB  (245). 
.'./.  CVD  is  measured  by  arc  AB  (Ax.  6). 
That  is,  /.  AVB  is  measured  by  arc  AB  (Ax.  6).         Q.E.D. 

727.  COR.  All  arcs  of  great  circles  containing  the  pole  of  a  great  cir- 
cle are  perpendicular  to  the  circumference  of  the  great  circle.     (See  540.) 

728.  COR.   A  spherical  angle  is  equal  to  the  plane  angle  of  the  dihe- 
dral angle  formed  by  the  planes  of  the  sides  of  the  angle.      (See  530.) 


729.     COR.   If  two  great  circles  are  perpendicular  to  each  other,  each 
of  their  circumferences  contains  the  pole  of  the  other.     (See  543;  702.) 


366 


SOLID   GEOMETRY 


730.  THEOREM.     A  sphere  may  be  inscribed  in  any  tetrahedron. 
Given  :  Tetrahedron  A-BCD. 

To  Prove:  (?). 

Proof  :  Pass  plane  OAB  bisecting  dih. 
Z  AB,  and  plane  OBC  bisecting  dih.  Z 
BC,  and  plane  OCD  bisecting  dih.  Z  CD, 
the  three  planes  meeting  at  point  O. 

Point  O,  in  plane  OAB,  is  equally  dis- 
tant from  faces  ABC  and  ABD  (?)  (551). 

Point  O,  in  plane  OBC,  is  equally  dis- 
tant from  faces  ABC  and  BCD  (?). 

Point  O,   in  plane  OCD,  is  equally  distant  from  faces  BCD 
and  ACD  (?). 

/.  O  is  equally  distant  from  all  four  faces  (Ax.  1). 

Hence,  using  O  as  a  center  and  the  perpendicular  distance 
OB  as  a  radius,  a  sphere  can  be  inscribed  (?)  (704).       Q.E.D. 

731.  COR.     The   six  planes  bisecting  the  six  dihedral  angles  of 
any  tetrahedron  meet  in  a  point. 

732.  THEOREM.    A  sphere  may  be  circumscribed  about  any  tetra- 
hedron. 

Given:  (?).     To  Prove:  (?). 

Proof  :  Take  E  and  F,  the  centers  of 
circles  circumscribed  about  the  faces 
ACD  and  BCD,  respectively.  Erect  EG, 
and  FH,  J_  to  these  faces.  Find  M,  the 
midpoint  of  edge  CD. 

EG  is  the  locus  of  points  equally 
distant  from  points  A,  D,  C  (?)  (526). 

FH  is  the  locus  of  points  equally 
distant  from  points  B,  C,  D  (?). 

That  is,  all  points  in  EG  and  FH  are  equally  distant  from  C 
and  D  (Ax.  1). 


BOOK  IX  367 

But  all  points  equally  distant  from  C  and  D  are  in  a  plane 
J.  to  CD  at  M  (?)  (525). 

.'.EG  and  FH  are  in  one  plane  and  are  not  parallel. 
(NotJ_  to  the  same  plane). 

That  is,  EG  and  FH  must  intersect  at  O. 

Hence,  O  is  equally  distant  from  A,  JS,  C,  and  D  (?)  (Ax.  1). 

That  is,  using  O  as  a  center  and  OA,  or  (XB,  or  OO,  or  OD,  as 
a  radius,  a  sphere  may  be  circumscribed  about  the  tetra- 
hedron A-BCD  (?)  (704).  Q.E.D. 

733.  COR.     Through  any  four  points  not  in  the  same  plane  a  sphere 
may  be  described. 

734.  COR.     The  six  planes  perpendicular  to  the  edges  of  any  tetra 
hedron  at  their  midpoints  meet  in  a  point.     (Explain.) 

735.  THEOREM.     The  intersection  of  two  spherical  surfaces  is  a 
circumference. 

Given  :  Two  intersecting  circum- 
ferences O  and  Or ;  common  chord 
CD ;  line  of  centers  XF,  intersecting 
CD  at  M. 

To  Prove  :  The  spherical  surfaces 
generated  by  the  revolution  of  these   (D,  intersect  in  a  cir- 
cumference. 

Proof:  If  these  ©  be  revolved  upon  XT  as  an  axis,  they 
will  generate  spheres  (?)  (714). 

CM=MD  (?)  (232). 

Point  (7,  common  to  both  CD,  will  generate  the  intersection 
of  the  spherical  surfaces  (?)  (482). 

CM  is  always  _L  to  XY  (?)  (232). 

.'.  the  curved  line  generated  by  C  is  in  one  plane  (?)  (502). 
.'.  the  intersection  is  a  circumference  (?)  (192).         Q.E.D. 


368 


SOLID   GEOMETRY 


CONSTRUCTIONS 
736.   PROBLEM.     To  find  the  radius  of  a  material  sphere. 
P  P' 


Given:  A  material  sphere.     Required:  To  find  its  radius. 

Construction :  First,  place  one  point  of  the  compasses  at  P, 
and  using  any  opening  of  the  compasses,  as  AP,  with  the 
other  point  draw  a  circumference  on  the  surface  of  the  sphere. 

Upon  this  circumference  take  three  points,  A  and  B  and  C, 
and  by  means  of  the  compasses  measure  the  straight  lines 
AB,  AC,  BC. 

Second,  construct  a  A  A'B'C',  whose  sides  are  AB,  AC,  BC. 
Circumscribe  a  circle  about  this  A,  and  draw  the  radius  AfDf. 

Third,  construct  a  right  A  P'A"D",  whose  hypotenuse  is 
the  known  line  PA  and  whose  leg  is  the  known  radius  ArDf. 
At  A"  erect  AnR!  _L  to  P'A"  meeting  P'D",  produced,  at  E' . 
Bisect  P'nf  at  Or. 


Statement :  orPr  =  the  required  radius. 


Q.E.F, 


Proof :  Points  P  and  o  are  equally  distant  from  the  points 
of  the  circumference  ABC  (Const,  and  707). 

.*.  P  is  in  the  line  _L  to  plane  ABC  at  the  center  D  (?) 
(526). 

If  this  J_  could  be  drawn  within  the  solid  sphere,  it  would 
be  a  diameter  (?) ;  arid  the  Z  PDA  would  be  a  rt.  Z  (?)  (489). 
AA'B'Cr  (?)  (58).  .'.DA  =  J)fA'  (227,201). 


BOOK   IX 


369 


Also,  A  PAD  =  A  P'A"D"  (?)  (73).     .'.Zp  =  Z  p'  (?) 
Z  P^t#  =  a  rt.  Z  (?).      .-.  A  PAR  =  A  P'A"R'  (?)  (74). 
.\PR  =  PfRf  (?).     Hence,  OP  =  o'p'   (Ax.  3). 

That  is,   O'P'  =  the  radius.  Q.E.D. 

737,   PROBLEM.   To  find  the  chord  of  a  quadrant  of  a  material  sphere. 

Given:  (?).     Required:  (?). 

Construction :  Find  the  radius  of  the 
sphere  (by  736).  Using  this  radius  OP 
and  any  center  O,  describe  a  semicircu in- 
ference PMR.  Erect  radius  OM  J_  to 
diameter  P#,  and  draw  PM. 

Statement:  Arc  PM  is  a  quadrant  of 
great  circle  of  the  given  sphere  and  chord 
PM  is  the  required  chord.  Q.E.F. 

Proof :  Arc  PM  is  a  quadrant  (?). 


; 


738.  PROBLEM.  To  describe  the  circumference  of  a  great  circle 
through  two  given  points  on  the  surface  of  a  sphere. 

Given :  The  points  A  and  B  on  the 
surface  of  the  sphere  O. 

Required:  (?). 

Construction :  Find  the  chord  of  a 
quadrant  of  the  given  sphere  (by  737). 

Place  one  point  of  the  compasses  at 
-4,  and  using  the  chord  just  found  as  an 
opening,  describe  an  arc  on  the  surface 
of  the  sphere. 

Similarly,  place  one  point  of  the  compasses  at  .B,  and 
using  the  same  opening,  describe  an  arc,  meeting  the  former 
arc  at  P. 

Now,  place  one  point  of  the  compasses  at  P  and  describe 
the  circumference  BAG,  using  the  chord  as  before. 

Statement:  (?).     Proof:  (Use  725.) 


370  SOLID   GEOMETRY 

739.  PROBLEM.     To  draw  an  arc  of  a  great  circle  through  a  given 
point  on  the  surface  of  a  sphere  and  perpendicular  to  a  given  great 
circle. 

Given :  Point  A  on  sphere  O,  and 
great  circle  BC,  whose  pole  is  P. 

Required :  To  draw  through  A  an 
arc  of  a  great  circle  J_  to  the  great 
circle  BC. 

Construction  :  Place  one  point  of  the 
compasses  at  A,  and  with  an  opening 
equal  to  the  chord  of  a  quadrant  of  the 

given  sphere  describe  an  arc  of  a  great  circle  intersecting 
the  given  great  circle  at  D. 

Now,  place  one  point  of  the  compasses  at  D  and  similarly 
draw  arc  of  great  circle  PAE.  Draw  PD,  the  arc  of  great 
circle  (by  738). 

Statement:  Arc  PAE  is  J_  to  circumference  BEDC.    Q.E.F. 

Proof:  ED,  EP,  and  PD  are  quadrants  (?)  (724). 

.-.  E  is  the  pole  of  arc  PD  (?)  (725). 

Hence,  Z.  PED  is  measured  by  quadrant  PD  (?)  (726). 

/.  ^  PED  is  a  right  angle  (247). 
That  is,  arc  PAE  is  _L  to  circumference  BEDC.     Q.E.D. 

SPHERICAL   TRIANGLES 

740.  A  spherical  triangle  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  arcs  of  great  circles. 

The  bounding  arcs  are  the  sides  of  the  triangle. 

The  intersections  of  the  sides  are  the  vertices  of  the 
triangle. 

The  angles  formed  by  the  sides  are  the  angles  of  the 
triangle. 

Spherical  triangles  are  equilateral,  equiangular,  isosceles, 
scalene,  acute,  right,  obtuse,  under  the  same  conditions  as  in 
plane  triangles. 


BOOK  ix 


371 


741.    A   birectangular    spherical    triangle    is    a    spherical 
triangle,  two  of  whose  angles  are  right  angles. 


A  trirectangular  spherical  triangle  is  a  spherical  triangle 
all  of  whose  angles  are  right  angles. 

The  unit  usually  employed  in  measuring  the  sides  of  a 
spherical  triangle  is  the  degree. 

It  is  obvious  that  the  circumferences  of  three  great 
circles  divide  the  surface  of  a  sphere  into  eight  spherical 
triangles. 

742.  Two  spherical  triangles  are  mutually  equilateral  if 
their  sides  are  equal  each  to  each;  and  they  are  mutually 
equiangular  if  their  angles  are  equal  each  to  each. 


MUTUALLY  EQUILATBRAL 

SPHERICAL  TRIANGLES 


MUTUALLY  EQUIANGULAR 
SPHERICAL  TRIANGLES 


POLAR 
TRIANGLES 


743.  If  three  great  circles  are  described,  having  as  their 
poles  the  vertices  of  a  spherical  triangle,  one  of  the  eight 
triangles  thus  formed  is  the  polar  triangle  of  the  first. 

The  polar  triangle  is  the  one  whose  vertices  are  nearest  the  vertices  of 
the  original  triangle. 


872 


SOLID  GEOMETRY 


744.  If   the    diameters   of   a   sphere   are 
drawn  to  the  vertices  of  a  spherical  triangle, 
the  original  triangle,  and  the  triangle  whose 
vertices  are  the  opposite  ends  of  these  diame- 
ters, are  symmetrical  spherical  triangles. 

745.  A  spherical  polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs  of  great  circles. 

Two  spherical  polygons  are  equal  if  they  can  be  made  to 
coincide.  The  diagonal  of  a  spherical  polygon  is  the  arc  of 
a  great  circle  connecting  two  vertices  not  in  the  same  side. 

Only  convex  spherical  polygons  are  considered  in  this  book. 


PRELIMINARY   THEOREMS 

746.  THEOREM.    The  planes  of  the  sides  of  a  spherical  triangle  form 
a  trihedral  angle : 

I.  Whose  vertex  is  the  center  of  the  sphere  (?)  (702). 
II.  Each  of  whose  face  angles  is  measured  by  the  intercepted  side 
of  the  triangle  (?)(245). 

III.  Each  of  whose  dihedral  angles  is  equal  to  the  corresponding 
angle  of  the  triangle  (?)  (728). 

747.  THEOREM.    A  pair  of  symmetrical  spherical  triangles  are  mutu- 
ally equilateral  and  mutually  equiangular.     (See  51,  206,  537,  728.) 

748.  THEOREM.    The  homologous  parts  of  a  pair 
of  symmetrical  spherical  triangles  are  arranged  in 
reverse  order. 

Proof :  If  the  eye  is  at  the  center  of  the 
sphere,  the  order  of  the  vertices  A,  B,  c  is  the 
same  in  direction  as  the  motion  of  the  hands 
of  a  clock.  But  the  order  of  Af,  Br,  Cf  is  in  the 
opposite  direction.  (See  556,  Note.)  Hence, 
the  parts  are  arranged  in  reverse  order.  Q.E.D. 


749.   THEOREM.    The  homologous  parts  of  two  symmetrical  spheri- 
cal triangles  are  equal.    (747.) 


BOOK   IX 


373 


750.  THEOREM.  Two  symmetrical  isos- 
celes spherical  triangles  can  be  superposed 
and  are  equal. 

Proof  :  The  method  of  superposition, 
as  in  the  case  of  plane  triangles. 


k751. 

.ngle  is 


THEOREMS  AND  DEMONSTRATIONS 


THEOREM.    One  side  of  a  spherical  tri- 
angle is  less  than  the  sum  of  the  other  two. 

Given:  (?).     To  Prove :  AB<AC  +  BC. 

Proof :  Draw  radii  OA,  OB,  OC. 

In  the  trihedral  Z  o,  Z  AOB  <  Z  AOC 

?)  (563). 

Z  AOB  is  measured  by  arc  AB,  etc.  (?). 
.*.  arc  AB  <  arc  AC  +  arc  BC  (Ax.  6). 


Q.E.D. 


752.  THEOREM.  In  a  birectangular  spherical  triangle  the  sides  op- 
posite the  right  angles  are  quadrants,  and  the  third  angle  is  measured 
by  the  third  side. 

Given:    Birectangular  A  ABC',   Z  B 
and  Z  C,  right  A. 

To  Prove  :  I.    AB  and  AC  quadrants. 

II.    Z  A  is  measured  by  arc  BC. 

Proof :  I.    Draw  radii  OA,  OB,  OC. 

Arc  AB  is  _L  to  arc  BC  and  arc  AC  is 
J_  to  arc  BC  (Hyp.). 

/.  A  is  the  pole  of  arc  BC  (?)  (729). 

/.  AB  and  ^4Gyare  quadrants  (?)  (724). 

II.   Z  A  is  measured  by  arc  BC  (?)  (726).  Q.E.D. 


753.   THEOREM.    If  two  sides  of  a  spherical  triangle  are  quadrants, 
the  triangle  is  birectangular. 

Proof :  A  is  the  pole  of  BC  (?)  (725). 

/.  A  B  and  c  are  rt.  A  (?)  (727).  Q.E.D. 


374 


SOLID   GEOMETRY 


754.  THEOREM.    The  three  sides  of  a  trirectangular  spherical  tri- 
angle are  quadrants. 

755.  THEOREM.    The  sum  of  the  sides  of  any  spherical  polygon  is 
less  than  360°. 

Given:  (?).     To  Prove:  (?). 

Proof :  Draw  radii  to  the  several 
vertices  of  the  polygon,  forming  the 
polyhedral  Z  O. 

Then,  Z  AOB  +  Z  BOG  +  Z  COD  + 
Z  AOD  <  360°  (?)  (564). 

But  Z  AOB  is  measured  by  arc  AB 
(?),  etc. 

.'.  arc  AB  +  arc  BC  +  arc  CD  +  arc 
AD  <  360°  (Ax.  6).  Q.E.D. 

756.  COR.    The  sum  of  the  sides  of  any  spherical  polygon  is  less 
than  the  circumference  of  a  great  circle. 

757. (  THEOREM.    If  one  spherical  triangle  is  the  polar  of  a  second 
triangle,  then  the  second  is  the  polar  of  the  first. 

Given :  Spherical  A  ABC  and  its 
polar  A  A'B'C'. 

To  Prove :  A  ABC  is  the  polar 
triangle  of  A  A'B'C'. 

Proof  :  A  is  the  pole  of  arc  BrCr 

(Hyp.)- 

.'.  B'  is  the  distance  of  a  quad- 
rant from  A  (?)  (724). 

C  is  the  pole  of  arc  AfBr  (?). 

.'.  B'  is  the  distance  of  a  quad- 
rant from  C  (?). 

Hence,  B'  is  the  pole  of  arc  AC  (?)  (725). 

Likewise,  A1  is  the  pole  of  BC,  and  c1  is  the  pole  of  AB. 

.'.  A  ABC  is  the  polar  A  of  A  A'B'C'  (?)  (743).          Q.E 


BOOK  IX 


3T5 


(758.  THEOREM.     In  two  polar  spherical  triangles  each  angle  of  one 
and  the  opposite  side  of  the  other  are  supplementary. 

Given:    Polar  A  ABC  and  A'B'C'. 

To  Prove: 

Z.A  +  a'  =  180°;  Z  A*  +  a  =  180°; 
Z  B  +  br  =  180°;  Z.B'  +  1  =  180°; 
Z  c  +  c'  =  180°;  Z  C'  +  c  =  180°. 

Proof:    Prolong  arc  B'C'  to  meet 
arc  AB  at  JR  and  arc  AC  at  5. 

JB'S  =  90°  and  c'fl  =  90°  (?)  (724). 
.•.B'flf+c'.R=1800(Ax.2). 


That  is,  Crs+Bfcr  +  C'R  or  ES  -{-B'C'  =  180°  (Ax.  4). 


Now,  JRS  is  the  measure  of  Z  J.  (?)  (726),  and  B'C  =  a. 

.-.Z.  A+a'  =  180°  (Ax.  6). 
Similarly,  Z  B  +  br  =  180°;  Z  c  +  c'  =  ISO0. 

Again,  prolong  arcs  AfBr  and  Arc'  to  meet  arc  BC  at  Jf 
and  L.     BL  =  90°  and  CJW  =  90°  (?)  (724). 

.-.  BL  +  CJf  =  180°  (Ax.  2). 

That  is,  LM  +MB  +  CM=  180°,  or  iJf  +  £C  =  180°  (Ax.  4). 
Now,  LM  is  the  measure  of  Z  Ar  (?),  and  _BC  =  a. 

.'./.  A'  +  a=  180°  (Ax.  6). 
Similarly,  Z  #'  +  &  =  180°;  Z  c'  +  c  =  180°.  Q.E.D. 

759.   THEOREM.     In  two  polar  spherical  triangles  each  angle  of  one 
is  measured  by  the  supplement  of  the  opposite  side  of  the  other. 
Proof  :    Identical  with  the  proof  of  758. 


Ex.  1.  What  is  the  locus  of  the  centers  of  those  spherical  surfaces 
that  pass  through  two  given  points? 

Ex.  2.  What  is  the  locus  of  the  centers  of  the  spherical  surfaces  of 
given  radius,  that  contain  two  given  points  ? 

Ex.  3.  W7hat  is  the  locus  of  the  centers  of  the  spherical  surfaces  that 
pass  through  three  given  points  ? 


376 


SOLID   GEOMETRY 


760.  THEOREM.  The  sum  of  the  angles  of  a 
spherical  triangle  is  greater  than  180°  and  less 
than  540°. 

Given:    A  spherical  A  ABC. 

To  Prove:    I.Z,4  +  Z£+Zc>  180°; 
II.  Z^t  +  Z£  +  Zc<  540°. 
Proof:    I.   Construct  A  A'B'C',  the  polar 
triangle  of  A  ABC. 


Z.A  +  a'  =  180°,  Z  B  +  bf  =  180°,  Z  c  +  a'  =  180°  (?)  (758). 
Adding,  Z^i  +  Zz?  +  ZC'+  a'  +  &'+<?'  =  540°  (Ax.  2). 
But,        _  a'  +  V  +  c'<  360°  (?)  (755). 


Subtracting,  Z^ 

II.  Again,  Z.  A  + 
But, 

Subtractin, 


+  Z.  C  +  a' 


>  180°  (Ax.  9). 

'  =  540°  (Ax.  2). 
'>      0°(?)(74Q). 


Z^  +  ZC  <  540°  (Ax.  9). 

Q.E.D. 

761.  COR.   The  sum  of  the  angles  of  a  spherical  triangle  is  greater 
than  two  right  angles  and  less  than  six  right  angles. 

762.  COR.   A  spherical  triangle  may  have  one,  two,  or  three  obtuse 
angles. 

V763.)  THEOREM.   Two  symmetrical  spherical  triangles  are  equiva- 
lent.^ 

Given:  Two  symmetrical  spheri- 
cal A  ABC  and  A'B'C'. 

To  Prove:    A  ABC  o  A  ArBfCr. 

Proof:  Suppose  p  is  the  pole  of 
the  circle  containing  A,  5,  C.  Draw 
the  diameters  AA  r,  BBf,  CC1,  pp', 
and  the  arcs  of  great  circles,  PA, 
PB,  PC,  P'A',  P'B',  P'C'. 

Z  POA  =  Z  P'OA'  (?). 

/.  arc  PA  =  arc  P'A'  (?)  (206). 


BOOK  IX  377 

Likewise,  arc  PB  =  arc  P'B'  and  arc  PC  =  arc  P!Cf. 
But  PA  =  PB  =  PC(?)  (723).    .'.p'Af  =  PfB'=p'c'  (Ax.  1). 
Hence,  A  APB  =  AAfPrBf  } 

A  APC  =  A  A'P'C'  |  (?)  (750). 


Adding,  A  ABC  ^  A  ArBrCf  (Ax.  2).  Q.E.D. 

NOTE.  If  the  pole  P  should  be  without  the  triangle  ABC,  one  of  the 
pairs  of  equal  triangles  would  be  without  the  original  triangles  and  would 
be  subtracted  from  the  sum  of  the  others  to  obtain  triangles  ABC  and 
A'B'C'. 

764.  THEOREM.  Provided  two  spherical  triangles  on  the  same 
sphere  (or  on  equal  spheres)  have  their  parts  arranged  in  the  same 
order,  they  are  equal: 

I.    If  two  sides  and  the  included  angle  of  one  are  equal  respec- 
tively to  two  sides  and  the  included  angle  of  the  other. 

II.   If  a  side  and  the  two  angles  adjoining  it  of  one  are  equal 
respectively  to  a  side  and  the  two  angles  adjoining  it  in  the  other. 

III.  If  three  sides  of  the  one  are  equal  respectively  to  three  sides 
of  the  other. 

Given:  (?).  A  R 

To  Prove:  (?). 

Proof:    I  and  II.     Superposition  as 
in  plane  triangles. 

III.  Draw  radii  of  the  sphere  to  all  the  vertices  of  the 
triangles. 

The  face  A  of  the  trih.  Z  O  =  the  face  A  of  the  trih.  Z  N, 
respectively.  (Explain.) 

Hence,  trih.  Z  o  =  trih.  Z  jy  (?)  (561). 

.*.  dih.  Z  o^  =  dih.  Z  NR-,   dih.  Z  o#  =  dih.  Z  NS;  etc. 

.'.  the  A  are  mutually  equiangular  (?)  (746,  III). 

Hence,  the  A  can  be  made  to  coincide. 

.'.  AABC=  A  BST  (?)  (28).  Q.E.D. 


378 


SOLID   GEOMETRY 


765.  THEOREM.    Provided   two   spherical   triangles    on  the   same 
sphere  (or  on  equal  spheres)  have  their  parts  arranged  in  reverse 
order,  they  are  symmetrical: 

I.   If  two  sides  and  the  included  angle  of  one  are  equal  respectively 
to  two  sides  and  the  included  angle  of  the  other. 

II.   If  a  side  and  the  two  angles  adjoining  it  of  one  are  equal  respec- 
tively to  a  side  and  the  two  angles  adjoining  it  of  the  other. 

III.  If  three  sides  of  one  are  equal  respectively  to  three  sides  of  the 
other. 

Proof:  In  each  of  these  cases 
construct  a  third  spherical  tri- 
angle, RrsrTr,  symmetrical  to 
the  A  EST. 

Then  A  R'S'T*  will  have  its 
parts  equal  to  the  parts  of  A 
ABC  and  arranged  in  the  same 
order.  (Explain.) 

.'.  A  E'S'T'  =  A  ABC  (?)  (764). 

Hence,  A  EST  is  symmetrical  to  A  ABC  (Ax.  6).       Q.E.D. 

766.  COR.  Two  mutually  equilateral  spherical  triangles  are  mutu- 
ally equiangular  and  are  equal  or  symmetrical. 

When  are  they  equal?    When  are  they  symmetrical? 

767.  THEOREM.    Two  mutually  equiangular  spherical  triangles  on 
the  same  sphere  (or  on  equal  spheres)  are  mutually  equilateral,  and 
are  equal  or  symmetrical. 

Given :  A  A  and  A',  mutually       /'*""\  X'""x 

equiangular. 

To  Prove  :  A  A  and  A'  mutu-  [ 
ally  equilateral,  and  equal  or  \ 
symmetrical. 

Proof :  Construct  A  E  and  E',  the  polar  A  of  A  and  Ar. 

The  sides  of  E  are  supplements  of  the  A  of  A     1  x9v 
and  the  sides  of  E'  are  supplements  of  the  A  of  A1  \ 


BOOK  IX 


379 


But  the  A  of  A  are  =  respectively  to  the  A  of  A1  (Hyp.). 

.'.  A  E  and  E'  are  mutually  equilateral  (?)  (49). 

Hence,  A  E  and  tf  are  mutually  equiangular  (?)  (766). 

Again,  A  A  and  A'  are  the  polar  A  of  E  and  ^  (?)  (757). 

.*.  the  sides  of  A  are  supplements  of  the  A  of  E  \  s^ 
and  the  sides  of  A1  are  supplements  of  the  A  of  E'  } 

Hence,  A  A  and  A'  are  mutually  equilateral  (?).     Also 
they  are  equal  (when?);    or  symmetrical  (when?).     Q.E.D. 

768.  THEOREM.   The  angles  opposite  the  equal  sides  of  an  isosceles 
spherical  triangle  are  equal. 

Given  :   (?).     To  Prove  :  Z  B  =  Z  c. 

Proof  :    Suppose  X  the  midpoint  of  BC. 

Draw  AX,  the  arc  of  a  great  circle. 

Now  the  two  spherical  A  ABX  and  ACX 
are  mutually  equilateral.     (Explain.) 

.'.  they  are  mutually  equiangular  and 
symmetrical  (?)  (766). 

=  Z  c(?)  (749). 

769.  COR.    The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  midpoint  of  the  base  bisects  the  vertex- 
angle  and  is  perpendicular  to  the  base.     (See  749.) 

770.  THEOREM.   If  two  angles  of  a  spherical  triangle  are  equal,  the 
sides  opposite  are  equal. 

Given:  (?).     To  Prove:  (?). 
Proof:    Construct  A  A'B'C',  the  polar 
triangle  of  A  ABC. 

Theii,.4/Ir  is  the  supplement  ofZ(7  j  .^ 


:C' 


and  A'  a'  is  the  supplement  of  Z  B 
/.  A'B'  =  A'C1  (?)  (49). 
Therefore,  Z  B'  =  Z.  Cf  (?)  (768). 

Again,  A  ABC  is  the  polar  triangle  of  A  A'B'C'  (?)  (757). 
And  AB  is  the  supplement  of  Z  c',  and  AC  of  Z  B'  (?). 

.'.AB  =  AC  (?).  Q.E.D 


380 


SOLID   GEOMETRY 


771.  THEOREM.  If  two  angles  of  a  spherical  triangle  are  unequal, 
the  sides  opposite  are  unequal  and  the  greater  side  is  opposite  the 
greater  angle. 

Given :  A  ABC  ;  Z  ABC  >  Z  C.  A 

To  Prove:  AC  >  AB. 


Proof :   Suppose  BR,  the  arc  of  a  great 
circle,  drawn,  making    Z  GBR  =  Z  C  and      ~. 
meeting  AC  at  R. 

Now  AR  +  BR  >  AB  (?)  (751).      But  BR  =  CR  (?)  (770). 

.'.  AR  +  CR  >  AB  (Ax.  6).      That  is,  AC  >  AB.         Q.E.D. 

772.  THEOREM.   If  two  sides  of  a  spherical  triangle  are  unequal,  the 
angles  opposite  are  unequal  and  the  greater  angle  is  opposite  the 
greater  side. 

To  Prove :  Z  ABC  >  Z  C. 

Proof  :    Z  ABC  is  either  <  Z  C  or  =  Z  C  or  >  Z  C. 

Continue  by  method  of  exclusion  (88). 

773.  THEOREM.     If  the  circumferences  of  two  circles  on  a  sphere 
contain  a  point  on  the  arc  of  a  great  circle  that  joins  their  poles,  they 
have  no  other  point  in  common. 

Given :  Point  P  on  the  arc  AB  of  a 
great  circle  of  a  sphere,  and  P  common 
to  two  circumferences  whose  poles  are  A 
and  B. 

To  Prove:  P  is  the  only  point  common 
to  these  circumferences. 

Proof:  Suppose  X  is  another  common 
point. 

Draw  arcs  of  great  CD,  AX  and  BX. 

Then  AX  +  BX  >  APB  (?)  (751). 
But  AX  =  AP    (?)  (723). 

Subtracting,  BX  >  BP    (?)  (Ax.  7). 

That  is,  X  is  without  the  O  B  and  cannot  be  in  both  the 
circumferences.  Q.E.D. 


BOOK  IX  381 


774.  COR.  If  two  circumferences  on  a  sphere  touch  each  other  at 
one  point,  and  only  one,  the  arc  of  a  great  circle  joining  their  poles  con- 
tains their  common  point. 

775.  THEOREM.  The  shortest  line  that  can  be  drawn  on  the  sur- 
face of  a  sphere,  between  two  points  on  the  surface,  is  the  less  arc  of 
the  great  circle  containing  the  two  points. 

Given:  Points  A  and  B,  and  AB 
the  arc  of  a  great  O  joining  them; 
line  ADEB,  any  other  line  on  the 
surface  of  the  sphere,  between  A 
and  B. 

To  Prove  :  Arc  AB  is  the  short- 
est line  on  the  surface,  that  can 
be  drawn  connecting  A  and  B. 

Proof :    Take   on  arc   AB   any 
point  C,  and  describe  two  circum- 
ferences through  c,  having  A  and  B  as  their  poles,  and  inter- 
secting ADEB  at  Z>  and  E.     Point  C  is  the  only  point  common 
to  these  two  ©  (?)  (773). 

No  matter  what  kind  of  line  AD  is,  a  line  of  equal  length 
can  be  drawn  from  A  to  C,  on  the  surface ;  and  a  line  can  be 
drawn  from  B  to  C  equal  in  length  to  BE. 

[Imagine ^4 D  revolved  on  the  surface  of  the  sphere,  using  A  as  a  pivot, 
and  D  will  move  along  the  circumference  to  coincide  with  point  C. 
Similarly  with  BE.'] 

There  is  now  a  line  from  A  to  JB,  through  C,  <  ADEB. 

That  is,  whatever  the  nature  of  ADEB,  there  is  a,  shorter 
line  from  A  to  -B,  which  contains  C,  any  point  of  arc  AB. 

Thus  the  shortest  line  contains  all  the  points  of  AB  and 
therefore  is  the  line  An.  Q.E.D. 

NOTE.  This  theorem  justifies  the  definition  of  the  "  distance  "  be- 
tween two  points,  etc.,  in  705. 


5  P 


\ 


382  SOLID  GEOMETRY 

776.  THEOREM.     Any  point  in  the  arc  of  a  great  circle  that  bisects 
a  spherical  angle  is  equally  distant  from  the  sides  of  the  angle. 

Given  :    Spherical  Z  EAG\  arc  AT  bisect- 
ing it  ;  any  point  P,  of  arc  AT\  PD  and  P#,  JjL 
arcs  of  great  (D  J_  to  AB  and  AC\  respec-           ..-*  /    >N 
tively. 

To  Prove  :    Arc  PD  =  arc  P^.  \ 

Proof :    The  arcs    PD    and  PE,   if    pro-  c    . 

longed,  will   pass   through  R   and    s,  re- 
spectively, the  poles  of  AB  and  AC  (?)   (729). 

Draw  arcs  AR  and  AS. 

Now,  Z  DAR  =  Z.EAS.     [Each  is  a  right  Z  ;  (727).] 
Z  D^P  =  Z  EAP  (Hyp.).     Subtracting, 
Z  RAP  =  Z  SAP  (Ax.  2). 

Also  RA  =  8A  (?)  (724$  and  AP  =  ^P. 

.-.A  R^P  is  symmetrical  to  A  SAP  (?)   (765,  I). 

.\RP  =  SP  (?).     But  ED  =  SE.  (Each  is  a  quadrant.) 

.*.  PD  =  PE  (Ax.  2).  Q.E.D. 

777.  THEOREM.     Any  point  on  the  surface  of  a  sphere  and  equally 
distant  from  the  sides  of  a  spherical  angle  is  in  the  arc  of  a  great  circle 
that  bisects  the  angle.    (The  proof  is  similar  to  the  proof  of  776.) 

ORIGINAL  EXERCISES 

1.  Vertical  spherical  angles  are  equal. 

2.  If  two  spherical  triangles,  on  the  same  or  equal  spheres,  are  mutu- 
ally equilateral,  their  polar  triangles  are  mutually  equiangular. 

3.  The  polar  triangle  of  an  isosceles  spherical  triangle  is  isosceles. 

4.  The  polar  triangle  of  a  birectangular  spherical  triangle  is  birec- 
tangular. 

5.  If  two  dihedral  angles  of  a  trihedral  angle  are  equal,  the  opposite 
face  angles  also  are  equal. 

Proof :  Construct  a  sphere  having  the  vertex  as  center,  etc. 

6.  If  two  face  angles  of  a  trihedral  a-ngle  are  equal,  the  opposite  di- 
hedral angles  also  are  equal. 


BOOK  IX  383 

7.  A  trirectangular  spherical  triangle  is  its  own  polar  triangle. 

8.  Two  symmetrical  spherical  polygons  are  equivalent. 

9.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of  the  other 
sides.     [Draw  diagonals  from  a  vertex.] 

10.  If  the  three  face  angles  of  a  trihedral  angle  are  equal,  the  three 
dihedral  angles  also  are  equal. 

11.  State  and  prove  the  converse  of  No.  10. 

12.  A  straight  line  cannot  meet  a  spherical  surface  in  more  than  two 
points. 

13.  If  two  dihedral  angles  of  a  trihedral  angle  are  unequal,  the  oppo- 
site face  angles  are  unequal,  and  the  greater  face  angle  is  opposite  the 
greater  dihedral  angle. 

14.  State  and  prove  the  converse  of  No.  13. 

15.  Two  lines  tangent  to  a  sphere  at  a  point  determine  a  plane  tan- 
gent to  a  sphere  at  the  same  point. 

16.  All  the  tangent  lines  drawn  to  a  sphere  from  an  external  point 
are  equal. 

17.  The  volume  of  any  tetrahedron  is  equal  to  one  third  the  product 
of  its  total  surface  by  the  radius  of  the  inscribed  sphere. 

18.  Every  point  in  the  circumference  of  a  great  circle  that  is  perpen- 
dicular to  an  arc  at  its  midpoint  is  equally  distant  from  the  ends  of  the 
arc. 

19.  The  points  of  contact  of  all  lines  tangent  to  a  sphere  from  an  ex- 
ternal point  lie  in  the  circumference  of  a  circle. 

20.  The  arcs  of  great  circles  perpendicular  to  the  sides  of  a  spherical 
triangle  at  their  midpoints  meet  in  a  point  equally  distant  from  the 
vertices. 

21.  If  the  opposite  sides  of  a  spherical  quadrilateral  are  equal,  the 
opposite  angles  are  equal. 

22.  If  the  opposite  sides  of  a  spherical  quadrilateral  are  equal,  the 
diagonals  bisect  each  other. 

23.  If  the  diagonals  of  a  spherical  quadrilateral  bisect  each  other,  the 
opposite  sides  are  equal. 

24.  The  exterior  angle  of  a  spherical  triangle  is  less  than  the  sum  of 
the  opposite  interior  angles. 

25.  The  sum  of  the  angles  of  a  spherical  quadrilateral  is  more  than 
four  right  angles  and  less  than  eight  right  angles. 


384  SOLID   GEOMETRY 

26.  If  two  spheres  are  tangent  to  each  other,  the  straight  line  joining 
their  centers  passes  through  the  point  of  contact. 

27.  The  sum  of  the  angles  of  a  spherical  polygon  is  more  than  2  n  —  4 
right  angles  and  less  than  2  n  right  angles. 

28.  The  arcs  of  great  circles  bisecting  the  angles  of  a  spherical  tri- 
angle meet  in  a  point. 

29.  A  circle  may  be  inscribed  in  any  spherical  triangle. 

30.  If  a  tangent  line  and  a  secant  be  drawn  to  a  sphere  from  an  ex- 
ternal point,  the  tangent  is   a  mean   proportional  between  the  whole 
secant  and  the  external  segment. 

31.  The  product  of  any  secant  that  can  be  drawn  to  a  sphere  from  an 
external  point,  by  its  external  segment,  is  constant  for  all  secants  drawn 
through  the  same  point. 

32.  If  two  spherical  surfaces  intersect  and  a  plane  be  passed  contain- 
ing their  intersection,  tangents  from  any  point  in  this  plane  to  the  two 
spherical  surfaces  are  equal. 

33.  Find  the  distance  from  the  center  of  a  sphere  whose  radius  is  15  to 
the  plane  of  a  small  circle  whose  radius  is  8. 

34.  The  polar  distance  of  a  small  circle  is  60°  and  the  radius  of  the 
sphere  is  12  in.     Find  the  radius  of  the  circle. 

35.  The  total  surface  of  a  tetrahedron  is  90  sq.  m.,  and  the  radius  of 
the  inscribed  sphere  is  4  m.     Find  the  volume  of  the  tetrahedron.      , 

36.  Find  the  radius  of  the  sphere  inscribed  in  a  tetrahedron  whose 
volume  is  250  and  total  surface  is  150. 

37.  Find  the  total  surface  of  a  tetrahedron  whose  volume  is  320,  if  the 
radius  of  the  inscribed  sphere  is  8. 

38.  Find  the  radius  of  the  sphere  inscribed  in  a  regular  tetrahedron 
whose  edges  are  each  10  in. 

39.  Find  the   radius  of  the  sphere  circumscribed   about  a  regular 
tetrahedron  whose  edges  are  each  18  in. 

40.  Find  the  radii  of  the  spheres  inscribed  in  and  circumscribed  about 
a  cube  whose  edges  are  each  10  in. 

41.  The  sides  of  a  spherical  triangle  are   60°,  80°,  110°.     Find  the 
angles  of  its  polar  triangle. 

42.  The  angles  of  a  spherical  triangle  are  74°,  119°,  87°.     Find  the 
sides  of  its  polar  triangle. 

43.  The  chord  of  the  polar  distance  of  the  circle  of  a  sphere  is  12,  and 
the  radius  of  the  sphere  is  9.     Find  the  radius  of  the  circle. 


BOOK  IX  385 

44.  The  polar  distance  of  a  circle  is  60°  and  the  diameter  of  the  circle 
is  8.     Find  the  diameter  of  the  sphere. 

[Denote  by,#,  each  side  of  an  equilateral  triangle  whose  altitude  is  4.] 

45.  The  radii  of  two  spherical  surfaces  are  11  in.  and  13  in.,  and  their 
centers  are  20  in.  apart.     Find  the  radius  of  the  circle  of  their  intersec- 
tion.    Find  also  the  distances  from  the  centers  of  the  spheres  to  the 
center  of  this  circle. 

46.  The  radii  of  two  spherical  surfaces  are  20  m.  and  37  in.,  and  the 
distance  between  their  centers  is  19  m.     What  is  the  length  of  the  diame- 
ter of  their  intersection  ? 

47     To  bisect  an  arc  of  a  great  circle. 

48.  To  draw  an  arc  of  a  great  circle  perpendicular  to  a  given  arc  of  a 
great  circle  through  a  given  point  in  the  arc. 

49.  To  bisect  a  spherical  angle. 

50.  To  bisect  an  arc  of  a  small  circle. 

51.  To  circumscribe  a  circle  about  a  given  spherical  triangle. 

52.  To  construct  a  spherical  angle  equal  to  a  given  spherical  angle  at 
a  given  point  on  the  same  sphere. 

53.  To  construct  a  spherical  triangle  having  the  three  sides  given. 

54.  To  construct  a  spherical  triangle  having  the  three  angles  given. 

55.  To  construct  a  plane  tangent  to  a  sphere  at  a  given  point  on  the 
surface. 

56.  To  construct  a  spherical  surface  having  the  radius  given  and  con- 
taining three  given  points. 

57.  To  construct  a  spherical  surface  that  shall  have  a  given  radius, 
touch  a  given  plane,  and  contain  two  given  points. 

58.  To  construct  a  spherical  surface  that  shall  have  a  given  radius, 
shall  be  tangent  to  a  given  sphere,  and  contain  two  given  points. 

59.  To  construct  a  spherical  surface  that  shall  contain  four  given 
points. 

60.  To  construct  a  plane  that  shall  contain  a  given  line  and  be  tan- 
gent to  a  given  sphere. 

61.  To  construct  a  plane  tangent  to  a  given  sphere  and  parallel  to  a 
given  plane. 

62.  What  is  the  locus  of  points  on  the  surface  of  a  sphere : 
(a)  Equally  distant  from  two  given  points  on  the  surface? 
(6)    Equally  distant  from  two  given  points  not  on  the  surface? 

(c)    Equally  distant  from  two  given  intersecting  arcs  of  great  circles  ? 

ROBBINS'    SOLID    GEOM. 25 


386  SOLID   GEOMETRY 

AREAS   AND   VOLUMES 

778.  A  lune  is  a  portion  of  the  surface  of  a  sphere  bounded 
by  two  semicircumferences  of  great  circles. 

The  points  of  intersection  of  the  sides  of  a  lune  are  the 
vertices  of  the  lune. 

The  angles  made  at  the  vertices  by  the  sides  are  the  angles 
of  the  lune. 


SPHERICAL    SPHERICAL       SPHERICAL    SECTOKS 
PYRAMID         SEGMENT  SPHERICAL 

CONE 


779.  A  zone  is  a  portion  of  the  surface  of  a  sphere  bounded 
by   the   circumferences    of    two    circles   whose    planes    are 
parallel. 

The  bases  of  a  zone  are  the  circumferences  bounding  it. 

The  altitude  of  a  zone  is  the  perpendicular  distance  between 
the  planes  of  its  bases. 

If  one  of  the  planes  is  tangent  to  the  sphere  the  zone  is  a 
zone  of  one  base. 

780.  A   spherical   degree   is   the   one    seven-hundred-and 
twentieth  part  of  the  surface  of  a  sphere.     If  the  surface  of 
a  sphere  is  divided  into  720  equal  parts,  each  part  is  a  spheri- 
cal degree. 

The  size  of  a  spherical  degree  depends  on  the  size  of  the  sphere. 
It  may  be  easily  conceived  to  be  half  a  lune  whose  angle  is  a  degree, 
that  is,  a  birectangular  spherical  triangle  whose  third  angle  is  1°. 
How  many  spherical  degrees  in  a  trirectangular  spherical  triangle  ? 


BOOK  IX  387 

781.  The  spherical    excess  of  a  spherical  triangle  is  the 
sum  of  its  angles  less  180°.     That  is,  E  =  A  +  B  +  C—  180°. 

782.  A  spherical  pyramid  is  a  portion  of  a  sphere  bounded 
by  a  spherical  polygon  and  the  planes  of  its  sides. 

The  vertex  of  a  spherical  pyramid  is  the  center  of  the 
sphere. 

The  base  of  a  spherical  pyramid  is  the  spherical  polygon. 

783.  A  spherical  sector  is  the  solid  generated  by  the  revo- 
lution of  the  sector  of  a  circle  about  any  diameter  of  the 
circle  as  an  axis. 

The  base  of  the  spherical  sector  is  the  zone  generated  by 
the  arc  of  the  circular  sector. 

A  spherical  cone  is  a  spherical  sector  whose  base  is  a  zone 
of  one  base. 

784.  A  spherical  segment  is  a  portion  of  a  sphere  included 
between  two  parallel  planes  that  intersect  the  sphere. 

The  bases  of  a  spherical  segment  are  the  circles  made  by 
the  parallel  planes. 

The  altitude  of  a  spherical  segment  is  the  perpendicular 
distance  between  the  bases. 

A  spherical  segment  of  one  base  is  a  segment,  one  of  whose 
bounding  planes  is  tangent  to  the  sphere. 

A  hemisphere  is  a  spherical  segment  of  one  base,  and  that 
base  is  a  great  circle. 

A  spherical  wedge  is  a  portion  of  a  sphere  bounded  by  a 
lune  and  the  planes  of  its  sides. 


Ex.  1.  What  is  the  spherical  excess  of  a  spherical  triangle  whose 
angles  are  60°,  70°,  and  100°? 

Ex.  2.    Distinguish  between  a  zone  and  a  spherical  segment. 

Ex.  3.  Find  the  area  of  a  spherical  degree  on  a  sphere  whose  surface 
is  3600  sq.  in. 

Ex.  4.  Find  the  area  of  a  spherical  triangle  containing  80  spherical 
degrees,  on  a  sphere  whose  surface  is  450  sq.  ft. 


388  SOLID  GEOMETRY 

PRELIMINARY   THEOREMS 

785.  THEOREM.    Either  angle  of  a  lune  is  measured  by  the  arc  of  a 
great  circle  described  with  the  vertex  of  the  lune  as  a  pole,  and  included 
between  the  sides  of  the  lune.     (See  726.) 

786.  COR.   The  angles  of  a  lune  are  equal. 

787.  THEOREM.     Every  great  circle  of  a  sphere  divides  the  sphere 
into  two  equal  hemispheres,  and  the  surface  into  two  equal  zones. 

788.  THEOREM.    The  spherical  excess  of  a  spherical  n-gon  is  equal 
to  the  sum  of  its  angles  less  (n  -  2)  180°. 

Proof  :  By  drawing  diagonals  from  any  vertex,  the  polygon 
is  divided  into  n  —  2  A. 

For  one&,E=S-  180°  (?)  (781). 

For  another  A,  Ef  =  8'  -  180°  (?).     Etc.  for  (n  -  2)  A. 

/.  by  adding,  the  excess  of  all  the  A  =  the  sura  of  all  their 
A-(n-V)  180°  (Ax.  2). 

That  is,  the  excess  of  a  spherical  polygon  =  the  sura  of  its 
//s_(n-2)  180°.  Q.B.D. 

789.  THEOREM.    If  a  regular  polygon  having  an  even  number  of  sides 
be  inscribed  in,  or  circumscribed  about,  a  circle,  and  the  figure  be  made  to 
revolve  about  one  of  the  longest  diagonals  of  the  polygon,  the  surface 
generated  by  the  polygon  will  be  composed  of  the  surface  of  cones, 
cylinders,  and  frustums,  and  the  surface  generated  by  the  circle  will  be 
a  spherical  surface. 

790.  THEOREM.    If  a  regular  polygon  having  an  even  number  of  sides 
be  inscribed  in,  or  circumscribed  about,  a  circle,  and  the  figure  be  made 
to  revolve  about  one  of  the  longest  diagonals  of  the  polygon,  the  surface 
generated  by  the  perimeter  of  the  polygon  will  approach  the  surface  of 
the  sphere  generated  by  the  circle,  as  a  limit,  if  the  number  of  sides 
of  the  polygon  is  indefinitely  increased, 

791.  THEOREM.    If  a  polyhedron  be  circumscribed  about  a  sphere 
and  the  number  of  its  faces  be  indefinitely  increased,  the  surface  of  the 
polyhedron  will  approach  the  surface  of  the  sphere  as  a  limit,  and  the 
volume  of  the  polyhedron  will  approach  the  volume  c-f  the  sphera  as  a 
limit. 


BOOK  IX  389 

THEOREMS   AND   DEMONSTRATIONS 

792.  THEOREM.  The  area  of  the  surface  generated  by  a  straight 
line  revolving  about  an  axis  in  its  plane  is  equal  to  the  product  of  the 
projection  of  the  line  upon  the  axis  by  the  circumference  of  a  circle 
whose  radius  is  the  line  perpendicular  to  the  revolving  line  at  its  mid- 
point, and  terminating  in  the  axis. 

Given  :  Line  AB  revolving  about  axis  XX! ; 
CD  —  projection  of  AB  on  XX]  \  M P  —  a  = 
JL  erected  at  midpoint  of  AB  and  terminating 
in  XX1 ';  MO  =  radius  of  midsection. 

To  Prove : 

Surface  generated  by  AB  =  CD  -  2  ?ra. 

Proof :  I.    The  surface  generated  by  AB  is 
the  surface  of  the  frustum  of  a  right  circu- 
lar cone  whose  bases  are  generated  by  AC  and  BD,  and  the 
midsection,  by  MO. 

Area  of  surface  =  2  TT  MO  •  AB  (696). 

Now,  AABH  and  MOP  are  similar  (?)  (321). 
.'.MO:  AH  =  MP:AB  (?). 

Hence,  MO  -  AB  =  An  -  MP  =  CD  •  a  (?). 

.*.  area  of  surface  —  2  irCD  -  a  —  CD  -  2  TTO,  (Ax.  6). 

II.  If  AB  is  II  to  XX1,  the  surface  is  cylindrical  and  equals 
CD'2-n-a  (?)  (665). 

III.  If  AB  meets  XX1  at  (7,  the  entire  surface  is  conical 
and  equals  irBD  -  AB  (?)  (693). 

Now  BD='2MO  (?)  (142)  ;  and  MO  •  AB  =  CD  •  a  (?). 

.'.  7T'BD'AB  =  7r-2  MO'AB  =  7r-2-CD-a  =  CD°  2  Tra  (Ax.  6). 

That  is,  the  area  of  the  surface  =  CD  •  2  ira  (Ax.  6).  Q.E.D. 


Ex.  1.  Find  the  spherical  excess  of  a  polygon  whose  angles  are  80°, 
110°,  140°,  130°,  160°. 

Ex.  2.  The  spherical  excess  of  a  spherical  polygon  is  the  difference 
between  the  sum  of  its  angles  and  the  sum  of  the  angles  of  a  plane  poly- 
gon having  the  same  number  of  sides. 


390 


SOLID   GEOMETRY 


793.     THEOREM.     The   surface  of  a  sphere  is  equivalent 
great  circles,  that  is,  to  4  irR2. 

Given  :  Semicircle  ACF\  diameter  AF-, 
S  =  surface  of  sphere  generated  by  revolv- 
ing the  semicircle  about  AF  as  an  axis ; 
H  =  radius  of  this  sphere. 

To  Prove  :   S  =  4  TrR2. 

Proof  :  Inscribe  in  this  semicircle  half  of 
a  regular  polygon  having  an  even  number 
of  sides.  Draw  the  apothems  and  denote 
them  by  a.  Draw  the  projections  of  the 
sides  of  the  polygon  on  the  diameter. 

Now,  if  the  figure  revolve  on  AF  as  an 
axis, 

the  surface  AB  =  AP  •  ZTTO, 
the  surface  BC  =  PS  -  2?ra 
the  surface  CD  =  ST  • 
etc.  etc. 


to  four 


(?)  (T92). 
Adding, 


the  entire  surface  =  (AP  +  PS  +  -ST  +  etc.)  -27ra  (Ax.  2). 

=  AF-Zira  (Ax.  6). 

Now  if  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  the  entire  surface  generated  by  the  polygon  will 
approach  S  (?)  (790),  and  a  will  approach  R  as  a  limit  (437). 
Also  AF  •  2  Tra  will  approach  AF'  2  TTR. 
.'.S  =  AF-27TR  (?)  (242).     But  AF=  %R  (?). 

.'.S=  4irB2  (Ax.  6).  Q.E.D. 


794.  THEOREM.  The  area  of  a  spherical  degree  equals  ^—  (780). 

795.  THEOREM.  The  areas  of  the  surfaces  of  two  spheres  are  to 
each  other  as  the  squares  of  their  radii  and  as  the  squares  of  their 
diameters.    (See  793.) 


Ex.  1.   What  is  the  area  of  the  surface  of  a  sphere  whose  radius  is 
10  in.  ?     What  is  the  area  of  a  spherical  degree  on  this  sphere  ? 


BOOK  IX 


391 


THEOREM.    The  area  of  a  zone  is  equal  to  the  product  of  its 
altitude  by  the  circumference  of  a  great  circle. 

Proof:  The  area  generated  by  chord  BC  (Fig.  of  793) 
=  PS-27ra  (?)  (792). 

If  the  number  of  sides  of  the  inscribed  polygon  is  indefi- 
nitely increased,  the  length  of  chord  BC  will  approach  arc  BC 
and  the  surface  generated  by  chord  BC  will  approach  the 
area  of  a  zone. 

Also,  PS  •  2  TTO,  will  approach  PS  •  2  TTR  (?). 

Hence,        area  of  zone  BC  =  PS-^irE  (?)  (242). 

If  the  altitude  of  the  zone  =  A, 

area  of  zone  =  2  irU  •  h.  Q.E.D. 


797.    THEOREM.    The  area  of  a  zone  of  one  base  is  equal  to 
of  a^circle  whose  radius  is  the  chord  of  the  generating  arc. 

Given:  Arc  AB  of  semicircle  ABC;  diameter 
AC;  chord  AB. 

To  Prove : 

Area  of  zone  generated  by  arc  AB  =  TT  A&*. 

Proof:  Area  of  zone  AB=AD-27rR  (796). 

That  is,  area  of  zone  AB  =  IT  -  AD  -2R. 

Draw  chord  BC.     A  ABC  is  a  rt.  A  (?). 

,'.AD'AC=AB2(i?)  (342). 

That  is,  AD  •  2  E  =  AB2  (Ax.  6). 

Hence,  area  of  zone  AB  =  TT  AB2    (Ax.  6). 

That  is,  area  of  zone  of  one  base  =  ir(chord)2. 


the  area 


Ex.  1.  On  a  sphere  whose  radius  is  6  in.,  find  the  area  of  a  zone  2|  in. 
in  height. 

Ex.  2.  What  does  the  formula  for  the  area  of  a  zone  become  when 
the  altitude  becomes  the  diameter  ? 

Ex.  3.  What  does  the  formula  for  the  area  of  a  zone  of  one  base  be- 
come when  the  generating  arc  becomes  a  semicircumference? 


392  SOLID   GEOMETRY 

798.  THEOREM.     The  area  of  a  lune  is  to  the  area  of  the  surface  of 
its  sphere  as  the  angle  of  the  lune  is  to  360? 

Given:  Lune  ABCDA  on  sphere  O; 
L  =  area  of  lune  ;  8  =  area  of  sphere  ; 
great  O  EB  whose  pole  is  A. 

To  Prove  :  L  :  S  =  Z  A  :  360. 

Proof:  I.  If  arc  BD  and  the  circum- 
ference of  O  EB  are  commensurable. 

There  exists  a  common  unit  of  meas- 
ure.    Suppose  this  unit  contained  5  times  in  BD ;  32  times  in 
the  circumference,     /.arc  BD  :  circumference  =  5  :  32  (?). 

That  is,  arc  BD  :  360  =  5  :  32.     Arc  BD  measures  Z  A  (726). 
.'./.A  :360  =  5  :32(Ax.  6). 

Pass  great  circles  through  the  several  points  of  division  of 
circumference  EB  and  vertex  A,  dividing  the  surface  of  the 
sphere  into  32  equal  lunes.     Then,    L  :  S  =  5  :  32  (Ax.  3). 
Hence,  L  :  8  =  Z  A  :  360  (Ax.  1).  Q.E.D. 

II.    If  the  arc  and  circumference  are  incommensurable. 
The  proof  is  similar  to  that  found  in  244,  302,  368,  539,  etc. 

799.  THEOREM.     The  number  of  spherical  degrees  in  the  area  of  a 
lune  is  double  the  number  of  degrees  in  its  angle. 

Proof :  Let  L°  denote  the  area  of  the  lune,  expressed  in 
spherical  degrees.     Then,  L°  :  S  =  Z  A  :  360  (?)  (798). 
That  is,  L°  :  720  =Z  A  :  360  (Ax.  6). 
Hence,  L°  =  2ZA.  Q.E.D. 


jj2 

800.   FORMULA.     The  area  of  a  lune  =  — —  x  z  A. 

Proof  :  L  :  s  =  Z  A  :  360  (?)  (798),  and  S  =  4  TTR*  (?). 

/.  L  :  4  7T722  =  Z  .4  :  360  (?).      .  •.  L  =  ^— -  x  Z  A.          Q.E.D. 

NOTE.     The  unit  of  measure  in  this  formula  is  the  square  unit. 


BOOK  IX 


801.  THEOREM.     Two  lunes  on  the  same  or  equal  spheres  are  to 
each  other  as  their  angles. 

Proof  :   L  :  s  =  Z  A  :  360,  and  L'  :  S  =  Z  A'  :  360  (?)  (798). 
Dividing,  L  :  L'  =  Z  A  :  Z  A'  (Ax.  3).  Q.E.D. 

802.  THEOREM.     Two  lunes   on  unequal    spheres,    having    equal 
angles,  are  to  each  other  as  the  squares  of  the  radii  of  the  spheres. 

Proof :   L  :  S  =  Z  A  :  360,  and  L1  :  s'  =  Z  A  :  360  (?). 
Hence,  L  :  S  =  L'  :  S'  (Ax.   1). 

.-.  L  :  L'=  S  :  S'  =  R*  :  E'*  (292  and  795). 

803.  THEOREM.     The  number  of  spherical  degrees  in  a  spherical 
triangle  is  equal  to  the  spherical  excess  of  the  triangle. 

Given:  Spherical  A  ABC  on 
sphere  O ;  spherical  excess  of 
the  A  =  E. 

To  Prove  :  Number  of  spher- 
ical degrees  in  A  ABC—E. 

Proof :  Continue  the  sides  of 
the  A  ABC  to  form  the  lunes 
ABA'CA,  BAB'CB,  CAC'BC; 
draw  diameters  AA\  BB1,  CC . 

A  ABC'  =0=  A  A'B'C  (?)  (763). 
Lune  CAC'BC  =  A  ABC  +  AAC'B  =c^  A  ABC  +  A  A'B'C  (Ax.  6). 
Now,  A  ABC  +  AA'B'Coluue  CAC'BC, 

Pand     A  ABC  +  A  A'BC=\ui\e  ABA'CA,    |  .(^x-  4). 
and     A  ABC  +  A  AB'c  =  lune  BAB'CB     j  Adding, 

2  A  ABC -{-A  ABC -{-A  A'B'C  +  AA'BC  +  A  AB'C 

=o=lune  A  +  lime  B  4-  lune  C  (Ax.  2). 
But,  A  ABC  +  AA'B'C  +  AA'BC  +  A  ^!B'c=360  spherical 
degrees  (?)  (713). 

Lune  A  +  lune  B  +  luneC  =  2Z^i  +  2ZB  +  2Zr7  (799). 

.'.  2  A  ABC  +360  =  2Z^  +  2ZJ5  +  2ZC  (Ax.  6). 
Hence,  A  ABC=Z  A  +  Z  £  +  Z  C-  180  =  # 

(Ax.  2,  Ax.  3,  781).     Q.E.D. 


394  SOLID   GEOMETRY 

804.  FORMULA.     The  area  of  a  spherical  triangle  =    ?2Q    x  E. 
Proof:  Area  of  one  spherical  degree  =    JT       (794). 
There  are  E  spherical  degrees  in  a  spherical  A  (?)  (803). 

/.the  area  of  a  spherical  triangle  =    ^Q   x  E.  Q.E.D. 

NOTES.     The  unit  of  measure  in  803  is  a  spherical  degree. 

The  unit  of  measure  in  804  is  a  square  unit  (sq.  in.,  sq.  ft.,  etc.). 

The  formula  of  804  reduces  to  ^f  • 

180 

The  area  of  a  spherical  triangle  is  determined  by  its  angles. 

805.  THEOREM.     The  number  of  spherical  degrees  in  a  spherical 
polygon  is  equal  to  its  spherical  excess. 

Given :   A  spherical  rc-gon. 

To  Prove:  The  number  of  spherical  de- 
grees in  this  w-gon  =  the  excess  of  the  poly- 
gon. 

Proof:  From  any  vertex  draw  diagonals,  dividing  the 
polygon  into  (n  —  2)  A ;  the  sums  of  the  A  of  these  A  are 
denoted  by  s,  sv  «2,  .  .  .  etc. 

Now,  the  number  of  sph.  degrees  in  one  A  =  s— 180°  (803); 
the  number  of  sph.  degrees  in  another  A  =  s^  — 180°  (?). 

Etc.,  for  (rc-2)  A. 

Adding,  the  number  of  sph.  degrees  in  the  w-gon 

=  the  sum  of  its  A-(n-  2)  180°  (Ax.  2). 

Excess  of  the  w-gon=  sum  of  its  z§-(rc-2)  180°  (788). 

.'.  the  number  of  spherical  degrees  in  a  spherical  polygon 

=  the  excess  of  the  polygon  (Ax.  1).  Q.E.D. 


Ex.  1.  Find  the  area  of  a  spherical  triangle  whose  angles  are  80°, 
125°,  and  95°,  on  a  sphere  whose  radius  is  6.3  in. 

Ex.  2.  Find  the  area  of  a  spherical  polygon  whose  angles  are  135°, 
105°,  85°,  155°,  120°,  on  a  sphere  whose  radius  is  15  ft. 


BOOK  IX 


806.  THEOREM.    The  volume  of  a  sphere  = 


395 


Given  :  Sphere  O  ;  radius  =  R  ;  surface  =  8  ;  volume  =  V. 

4   7TE3 

To  Prove  :   V  =  —  «  —  • 
o 

Proof:  Suppose  a  polyhedron  cir- 
cumscribed about  the  sphere,  its 
surface  denoted  by  81  and  its  vol- 
ume by  V1  .  Suppose  planes  be 
passed  through  the  edges  of  the 
polyhedron  and  the  center  of  the 
sphere,  thus  dividing  the  polyhe- 
dron into  pyramids  whose  vertices 
are  all  at  the  center,  and  whose  common  altitude  is  R. 

The  volume  of  one  such  pyramid  =  j  R  -  its  base  (?)  (625). 

.*.  volume  of  all  the  pyramids  =  J  R  -  sum  of  all  their  bases 
(Ax.  2)  ;  that  is,  v'  =  J  R  •  S1  . 

Indefinitely  increase  the  number  of  faces  of  the  polyhe- 
dron, thus  indefinitely  decreasing  each  face, 

and  vr  will  approach  Fas  a  limit)  ,?,   fTQI^ 
and  sr  will  approach  S  as  a  limit  j 

Hence,  ^  R  -  Sr  will  approach  i  R  •  S  as  a  limit  (?). 

Therefore,  F=  J  R  -  S  (?)  (242).    But  s  =  4  TTR  2  (?). 


:.V 


8 


6). 


Q.E.D. 


807.   THEOREM.    The  volumes  of  two  spheres  are  to  each  other  as 
the  cubes  of  their  radii  or  as  the  cubes  of  their  diameters. 

Proof : 


;,.£-..      (Explain.) 

V  '6  -6  X"       ^JJ')»        D'3  QED 

808.  THEOREM.  The  volume  of  a  spherical  pyramid  is  equal  to  one 
third  the  product  of  the  polygon  that  is  its  base,  by  the  radius  of  the 
sphere. 

Proof :  Similar  to  the  proof  of  80C. 


396 


SOLID   GEOMETRY 


809.   THEOREM.    The  volume  of  a  spherical  wedge  is  to  the  volume 
of  the  sphere  as  the  angle  of  its  base  is  to  360. 

Proof:   Similar  to  the  proof  of  798. 


810.  THEOREM.    The  volume  of  a  spherical  sector  is  equal  to  one 
third  the  product  of  the  zone  that  is  its  base  by  the  radius  of  the  sphere. 

Proof  :  Similar  to  the  proof  of  806. 

811.  FORMULAS.   Vol.  of  a  spherical  sector  =  J  R-  zone  (810). 
But  the  zone  —  ^TrR-h  (?).     Therefore, 

1.  The  volume  af  spherical  sector  =  i  irK2  •  h  (Ax.  6). 

3 

2.  The  volume  of  a  spherical  cone  =  f  irJB2  •  h  (811,  1). 

8 


3.    The  volume  of  a  spherical  wedge  = 


(from  809). 


812.   PROBLEM.     To  find  the  volume  of  a  spherical  segment. 


Ipherical  segment  of  one  base. 
Given :  Spherical  segment  generated  by 
the  figure  ACX\  semicircle  XAY\  AC  =  r j 
radius  of  sphere  =  R ;  altitude  =  CX  =  h. 

Required :    To   find   the   volume   of    the 
spherical  segment. 

Computation:  Draw  chords  AX,  AY,  and 
radius  AO. 

The  right  A  ACO  will  generate  a  cone  of 
revolution  (?)  (671). 

The  volume  of  spherical  segment  A CX  =  volume  of  spher- 
ical cone  OA  X  minus  volume  of  cone  A  CO. 

Volume  of  spherical  cone  OAX  =  |  TrK2  •  h  (?)  (811,  2)  ; 

volume  of  cone  ACO  =  J  vrr2  -CO  (?)  (695). 
Now  r2=  CX-CY=h  (2  s-h)  (?)  (340,11);   and 
CO  =  R  -  h.    '.'.  vol.  ACO  =  J-  Trh  (2  R  -  h)  (R  -  h)  (Ax.  6). 
Hence,  volume  of  spherical  segment  ACX 
=  f  7rR2h  -  (f  77 R2h  -  TrRh2  +  J  7r£3)   (Ax.  6).     That  is, 
volume  of  spherical  segment  of  one  base  =  j  ivh2  (3  R  -  h}. 


^ 


BOOK  IX 


397 


2.  Spherical  segment  not  including  the 
center. 

Given  :  Spherical  segment  generated  by 
figure  ACDB:,  semicircle  XAEY\  AG  —  r\ 
BD  =  r' ;  radius  of  sphere  =  R ;  altitude  = 

Required:  To  find  the  volume  of  the 
spherical  segment. 

Computation  :  The  A  AGO  and  BDO  gen- 
erate cones  of  revolution  (?)  (671). 

The  volume  of  spherical  segment  ACDB 

=  volume  of  spherical  sector  ABO 
plus  the  volume  of  cone  A  CO 
minus  the  volume  of  cone  BDO. 

Each  of  these  volumes  can  be  determined  from  formulas 
already  established. 

3.  Spherical  segment  including  the  center. 

Given  :   Spherical  segment  generated  by  figure  BDSR  ;  etc. 

Required  :   (?).     Computation  :  The  same  as  that  of  case  2, 
except  that  both  cones,  BDO  and  RSO,  dreUdded. 

II 

ORIGINAL  EXERCISES 

1.  Prove  that  the  area  of   the  surface  of  a  sphere   is  equal  to  the 
square  of  the  diameter  multiplied  by  TT,  that  is,  S  =  7rZ>2. 

2.  Prove  that  the  volume  of  a  sphere  is  equal  to  one  sixth  the  cube 
of  the  diameter  multiplied  by  TT,  that  is,  V  =  \  TrZ)3. 

3.  The  surface  of  a  sphere  is  equal  to  the  cylindrical  surface  of  the 
circumscribed  cylinder. 

4.  The  total  surface  of  a  hemisphere  is  three  fourths  the  surface  of 
the  sphere. 

5.  The  volume  of  a  sphere  is  two  thirds  the  volume  of  the  circum- 
scribed cylinder. 


398 


SOLID   GEOMETRY 


6.  Upon  the  same  circle  as  a  base  are  constructed  a  hemisphere,  a 
cylinder  of   revolution,  and  a  cone  of  revolution,  all  having  the  same 
altitude.     Prove  that  their  total  areas  are  3w/22,  lirR*,  TrR\l  +  V2), 
respectively,  and  their  volumes  are  f  TrR8,  TrR8,  ^  TrR3,  respectively. 

7.  Two  zones  on  the  same  sphere,  or  on  equal  spheres,  are  to  each 
other  as  their  altitudes. 

8.  The  area  of  the  surface  of  a  sphere  is  equal  to  the  area  of  the 
circle  whose  radius  is  the  diameter  of  the  sphere. 

9.  Show  that  the  formula  for  the  volume  of  a  spherical  segment  of 
one  base  reduces  to  the  correct  formula  for  the  volume  of  a  hemisphere, 
when  the  base  of  the  segment  is  a  great   circle ;   and  to  the   correct 
formula  for  the  volume  of  a  sphere  when  the  planes  are  both  tangent. 

10.  In  an  equilateral  triangle  is  inscribed  a  circle, 
and  the  figure  is  revolved  about  an  altitude  of  the 
triangle  as  an  axis.     Prove, 

(a)  That  the  surface  generated  by  the  circumfer- 
ence is  two  thirds  the  lateral  surface  generated  by  the 
triangle. 

(6)  That  the  volume  generated  by  the  circle  is  four 
ninths  the  volume  generated  by  the  triangle. 

11.  Derive  a  formula  for  the  surface  of  a  sphere,  containing  only  V 
andir. 

12.  Derive  a  formyja  for  the  volume  of  a  sphere,  containing  only  S 
andir. 

13.  In    a    circle  whose    radius  is    R,   there    are 
inscribed  a  square  and  an  equilateral  triangle  having 
their  bases  parallel  ;   the  whole   figure    is    then  re- 
volved about  the  diameter  perpendicular  to  the  base 
of  the  triangle.     Find,  in  terms  of  R, 

(a)  The  total  areas  of  the  three  surfaces  generated; 
(6)  The  volumes  of  the  three  solids  generated. 

14.  If  a  cylinder  of  revolution  having  its  altitude  equal  to  the  diam- 
eter of  its  base,  and  a  cone  of  revolution  having  its  slant  height  equal  to 
the  diameter  of  its  base  are  both  inscribed  in  a  sphere, 

(a)  The  total  area  of  the  cylinder  is  a  mean  proportional  between  the 
area  of  the  surface  of  the  sphere  and  the  total  area  of  the  cone; 

(6)  The  volume  of  the  cylinder  is  a  mean  proportional  between  the 
volume  of  the  sphere  and  the  volume  of  the  cone. 


BOOK  IX  399 

15.  About  a  circle  whose  radius  is  R,  there  are  circumscribed  a  square 
and  an  equilateral  triangle  having  their  bases  in  the 

same  straight  line.  The  whole  figure  is  then  re- 
volved about  an  altitude  of  the  triangle.  Find,  in 
terms  of  R, 

(a)  The  total  areas  of  the  three  surfaces  gen- 
erated. 

(b)  The  volumes  of  the  three  surfaces  generated. 

16.  If  a  cylinder  of  revolution  having  its  altitude  equal  to  the  diam- 
eter of  its  base,  and  a  cone  of  revolution  having  its  slant  height  equal 
to  the  diameter  of  its  base,  be  circumscribed  about  a  sphere, 

(a)  The  total  area  of  the  cylinder  is  a  mean  proportional  between  the 
area  of  the  surface  of  the  sphere  and  the  total  area  of  the  cone  ; 

(ft)  The  volume  of  the  cylinder  is  a  mean  proportional  between  the 
volume  of  the  sphere  and  the  volume  of  the  cone. 

17.  The  line  joining  the  centers  of  two  intersecting  spherical  sur- 
faces is  perpendicular  to  the  plane  of  the  intersection  at  the  center  of 
the  intersection. 

18.  A  cube  and  a  sphere  have  equal  surfaces;  show  that  the  sphere 
has  the  greater  volume. 

19.  Prove  that  the  parallel  of  latitude  through  a  point  having  30° 
north  latitude  bisects  the  surface  of  the  northern  hemisphere. 

20.  Prove  that  in  order  that  the  eye  may  observe 
one  sixth  of  the  surface  of  a  sphere,  it  must  be  at  a 
distance  from  the  center  of  the  sphere  equal  to  f  of 
the  radius. 

Proof:     Zone   TT  =  $   surface   of   sphere   (Hyp.). 
:.AB  =  £  diam.^i  /?.     Hence,  EC  =  %R. 

In  rt.  A  ETC,  TC2  =  EC-BC  (?)  ;   .*.  R*  =  EC  .  f  R,  or  EC  =  f  R 
(Explain.)  Q.E.D. 

21.  How  many  miles  above  the  surface  of  the  earth  (diameter  of 
earth  =  7960  mi.)  must  a  person  be  in  order  that  he  may  see  one  sixth  of 
the  earth's  surface  ? 

22.  If  the  area  of  a  zone  of  one  base  is  a  mean  proportional  between 
the  area  of  the  remaining  zone  of  the  sphere  and  the  area  of  the  entire 
sphere,  the  altitude  of  the  zone  is  R(  Vo  —  1). 

23.  The  area  of  a  lune  is  to  the  area  of  a  trirectangular  spherical 
triangle  as  the  angle  of  the  lune  is  to  45°. 

24.  A  cone,  a  sphere,  and  a  cylinder  have  the  same  diameters  and 
altitudes.     Prove  that  their  volumes  are  in  arithmetical  progression. 


• 

400  SOLID   GEOMETRY 

25.  The  surface  of  a  sphere  bears  the  same  ratio  to  the  total  surface 
of  the  circumscribed  cylinder  of  revolution,  as  the  volume  of  the  sphere 
bears  to  the  volume  of  the  cylinder. 

26.  Ttee  smallest  circle  upon  a  sphere,  whose  plane  passes  through  a 
given  point  within  the  sphere,  is  the  circle  whose  plane  is  perpendicular 
to  the  diameter  through  the  given  point. 

27.  What  part  of  the  surface  of  the  earth  could  one  see  if  he  were 
at  the  distance  of  a  diameter  above  the  surface  ? 

28.  Prove  that  if  any  number  of  lines  in  space  be  drawn  through  a 
point,  and  from  any  other  point  perpendiculars  to  these  lines  be  drawn, 
the  feet  of  all  of  these  perpendiculars  lie  on  the  surface  of  a  sphere. 

29.  The  volume  of  a  sphere  is  to  the  volume  of  the  circumscribed 
cube  as  TT  :  6.     The  volume  of  a  sphere  is  to  the  volume  of  the  inscribed 
cube  as  TT  :  |  V3. 

30.  There  are  five  spheres  that  touch  the  four  planes  of  the  faces  of 
a  tetrahedron. 

31.  If  two  angles  of  a  spherical  triangle  are  supplementary,  the  sides 
of  the  polar  triangle,  opposite  these  angles,  are  supplementary. 

32.  A  square,  whose  side  is  a,  is  revolved  about  a  diagonal,  and  also 
about  an  axis  bisecting  two  opposite  sides.     Which  of  these  figures  con- 
tains the  greater  volume  ?    Which  has  the  greater  surface  ? 


33.  Find  the  area  of  the  surface,  and  the  volume  of  a  sphere  whose 
radius  is  6. 

34.  Find  the  area  of  a  zone  whose  altitude  is  4  on  a  sphere  whose 
radius  is  14. 

35.  Find  the  area  of  a  lune  whose  angle  is  30°  on  a  sphere  whose 
radius  is  8  in. 

36.  Find  the  area  of  a  spherical  triangle  whose  angles  are  110°,  41°, 
92°,  on  a  sphere  whose  radius  is  10. 

37.  Find  the  volume  of  a  sphere  whose  radius  is  5. 

38.  Find  the  volume  of  a  spherical  pyramid  whose  base  is  35  sq.  in., 
on  SL sphere  whose  radius  is  12  in. 

39.  Find  the  area  of  a  spherical  polygon  whose  angles  are  87°,  108°, 
121°,  128°,  on  a  sphere  whose  radius  is  25. 

40.  What  is  the  radius  of  a  sphere  whose  surface  is  1386  sq.  yd.  ? 

43,,.  What  is  the  radius  of  a  sphere  whose  volume  is  — - —  cu.  in.  ? 


BOOK  IX  401 

42.  What  is  the  area  of  the   surface  of  a  sphere  whose  volume  is 
288  TT  cu.  ft.  ? 

43.  What  is  the  volume  of  a  sphere,  the  area  of  whose  surface  is  2464 
sq.  in.  ? 

44.  Find  the  area  of  a  zone  whose  altitude  is  3J,  if  the  radius  of  the 

sphere  is  7|. 

45.  Find  the  volume  of  a  spherical  sector  the  altitude  of  whose  base 
is  5£  in.  if  the  radius  of  the  sphere  is  6  in. 

46.  Find  the  diameter,  the  circumference  of  a  great  circle,  and  the 
volume  of  a  sphere  the  area  of  whose  surface  is  25  TT  sq.  ft. 

47.  By  how  many  cubic  inches  is  a  9-in.  cube  greater  than  a  9-in. 
sphere  ? 

48.  The  radius  of  a  sphere  is  15,  and  the  angles  of  the  base  of  a 
spherical  pyramid  are  160°,  127°,  96°,  145°,  and  117°.     Find  the  volume 
of  the  pyramid. 

49.  A  cylindrical  vessel  10  in.  in  diameter  contains  a  liquid.     A  metal 
ball  is  immersed  in  the  liquid  and  the  surface  rises  £  in.     What  is  the 
diameter  of  the  ball? 

50.  If  a  sphere  3  ft.  in  diameter  weighs  99  Ibs.,  what  will  a  sphere  of 
the  same  material  4  ft.  in  diameter  weigh  ? 

51.  The  radii  of  the  bases  of  a  frustum  of  a  cone  of  revolution  are 
5  in.  and  6  in.,  and  the  altitude  of  the  frustum  is  19£  in.     What  is  the 
diameter  of  an  equivalent  sphere  ? 

52.  What  is  the  radius  of  a  sphere  whose  surface  is  equivalent  to  the 
total  surface  of  a  right  circular  cylinder  having  an  altitude  equal  to  21, 
and  radius  of  the  base  equal  to  6  V 

53.  Find  the  volume  generated  by  the  revolution  of  an  equilateral 
triangle  inscribed  in  a  circle  whose  radius  is  8,  about  an  altitude  of  the 
triangle  as  an  axis.     (See  Fig.  of  Ex.  55.) 

54.  In  the  figure  of  No.  55,  find  the  volume  of  the  segment  generated 
by  the  figure  A  ED  revolving  about  CD  as  an  axis. 

55.  Find  the  area  of  the  surface,  and  the  volume 
of  the  sphere  generated  by  a  circle  that  is  circum- 
scribed about  an  equilateral  triangle  whose  side  is  10. 

56.  Circumscribing  a  sphere  whose  radius  is  18,  is 
a  cylinder  of  revolution.     Compare  their  total  areas. 
Their  volumes. 

BOBBINS'  SOLID  GEOM, — 26 


402  SOLID   GEOMETRY 

57.  Circumscribing  a  cylinder  of  revolution  whose 
altitude  and  diameter  are  each  6  in.,  is  a  sphere.     Find 
the  volume  and  area  of  the  surface  of  the  sphere. 

58.  Circumscribing  a  cylinder  whose   altitude  is  4 
and  diameter  is  3,  is  a  sphere.     Find  the  radius  and 
volume  of  the  sphere. 

59.  Each  edge  of  a  cube  is  8  in.    What  is  the  area  of  the  surface,  and 
the  volume  of  the  circumscribed  sphere  ? 

60.  Find  the  volume  of  one  of  the  segments  cut  from  a  10  in.  sphere 
by  the  plane  of  one  of  the  faces  of  the  inscribed  cube. 

61.  The  volume  of  a  certain  sphere  is  179f  cu.  ft.     Find  the  radius 
of  a  sphere  8  times  as  large.     Find  the  radius  of  a  sphere  3  times  as 
large. 

62.  The  radius  of  a  certain  sphere  is  5  in.     What  is  the  radius  of  a 
sphere  twice  as  great ?     Half  as  great?    Two  thirds  as  great ? 

63.  A  hollow  sphere  has  an  outer  diameter  of  20  in.,  and  an  inner 
diameter  of  16  in.     Find  the  volume  of  the  metal  in  the  shell. 

64.  Find  the  diameter  of  that  sphere  whose  volume  is,  numerically, 
equal  to  the  area  of  its  surface. 

65.  A  projectile  consists  of  a  right  circular  cylinder  having  a  hemi- 
sphere at  each  end.     If  the  cylinder  is  9  in.  long  and  7  in.  in  diameter, 
what  is  the  volume  of  one  projectile  ? 

66.  Inscribed  in  a  regular  tetrahedron  whose  edge  is  4,  and  circum- 
scribed about  it  are  two  spheres.     Find  their  radii. 

67.  Find  the  radii  of  the   spheres  inscribed  in  and  circumscribed 
about  a  regular  hexahedron  whose  edge  is  8  m. 

68.  Find  the  radii  of  the  spheres  inscribed  in  and  circumscribed  about 
a  regular  octahedron  whose  edge  is  12  in. 

69.  How  many  spherical  bullets  |  in.  in  diameter  can  be  made  from 
a  cube  of  lead  5  inches  on  each  edge  ? 

70.  The  area  of  a  spherical  triangle  whose  angles  are  158°,  77°,  95°,  is 
288|.     Find  the  radius  of  the  sphere. 

71.  The  area  of  a  spherical  triangle  whose  excess  is  75,  is  135  TT. 
Find  the  radius  of  the  sphere. 

72.  If  the  radius  of  a  sphere  is  2.5,  and  the  sides  of  a  triangle  on  it 
are  104°,  115°,  101°,  find  the  area  of  the  polar  triangle. 


BOOK  IX  403 

73.  In  a  trihedral  angle  the  plane  angles  of  the  dihedral  angles  are 
75°,  85°,  110°.     Find  the  number  of  degrees  of  surface  of  a  sphere,  whose 
center  is  the  vertex  of  the  trihedral  angle,  inclosed  by  the  faces  of  this 
trihedral  angle. 

74.  What  is  the  area  of  a  spherical  hexagon,  each  of  whose  angles  is 
145°,  on  a  sphere  whose  radius  is  15? 

75.  How  many  miles  above  the  earth  would  a  person  have  to  be  in 
order  that  he  may  see  a  third  of  its  surface  ?    One  eighth  of  its  surface? 

76.  Find  the  altitude  of  the  zone  whose  area  is  equal  to  the  area  of 
a  great  circle  of  a  sphere. 

77.  If  the  radius  of  a  sphere  is  doubled,  how  is  the  amount  of  surface 
affected?    The  volume?    The  weight? 

78.  At  a  distance  (=  d)  from  the  center  of  a  sphere  whose  radius  is 
r,  is  an  illuminating  point.     What  is  the  altitude  of  the  zone  illuminated? 

79.  On  a  sphere  having  a  radius  of  5  in.  is  an  equiangular  spherical 
triangle  whose  area  is  5  TT  sq.  in.     Find  the  angles  of  the  triangle. 

80.  Find  the  area  of  the  surface  of  a  sphere  whose  volume  is  a  cu.  yd. 

81.  Find  the  volume  of  a  sphere  whose  surface  is  a  sq.  yd. 

82.  If  a  circumference  is  described  on  the  surface  of  a  sphere,  by  a 
pair  of  compasses  whose  points  are  2f  in.  apart,  what  is  the  area  of  the 
zone  bounded  by  this  circumference  ? 

83.  On  a  sphere  the  area  of  whose  surface  is  288  sq.  ft.  is  a  birectan- 
gular  spherical  triangle  whose  vertex  angle  is  100°.     Find  the  area  of 
this  triangle. 

84.  Five  inches  from  the  center  of  a  sphere  whose  diameter  is  two 
feet,  a  plane  is  passed.     Find  the  areas  of  the  two  zones  formed.     Find 
the  chords  of  their  generating  arcs. 

85.  The  diameter  of  the  moon  is  about  2000  mi. ;  that  of  the  earth, 
about  8000  mi.     How  do  their  surfaces  compare?     Their  volumes? 

86.  The  radii  of  two  concentric  spheres  are  12  and  13  in.     A  plane 
is  tangent  to  the  inner  sphere.    Find  area  of  section  of  outer  sphere. 

87.  If  a  solid  sphere  4  ft.  in  diameter  weighs  500  Ibs.,  what  is  the 
weight  of  a  spherical  shell  whose  external  diameter  is  10  ft,  made  of  the 
same  material  and  a  foot  thick? 

88.  The  sun's  diameter  is  about  109  times  the  diameter  of  the  earth. 
How  do  the  areas  of  their  surfaces  compare?    Their  volumes? 

89.  How  many  quarter-inch  spherical  bullets  can  be  made  from  a 
sphere  of  lead  a  foot  in  diameter  ? 


404  SOLID   GEOMETRY 

90.  The  angles  of  a  spherical  triangle  are  80°,  90°,  100°.    Find  the  angle 
of  an  equivalent  lune. 

91.  Find  the  angles  of  an  equiangular  spherical  triangle  equivalent 
to  the   sum  of  three  equiangular  spherical  triangles  (upon  the  same 
sphere)  whose  angles  are  each  75°. 

92.  What  is  the  radius  of  a  sphere  equivalent  to  the  sum  of  two 
spheres  whose  radii  are  3  in.  and  4  in.,  respectively? 

93.  What  is  the  radius  of  a  sphere  equivalent  to  the  difference  of 
two  spheres  whose  radii  are  5  in.  and  4  in.,  respectively? 

94.  The  area  of  an  equiangular  spherical  triangle  is  TT,  and  the  radius 
of  the  sphere  is  4.     Find  the  angles  of  the  triangle. 

95.  The  volumes  of  two  spheres  are  to  each  other  as  64  :  343.     What 
is  the  ratio  of  their  surfaces? 

96.  Find  the  volumes  of  the  segments  of  a  sphere  whose  radius  is  12, 
formed  by  a  plane  whose  distance  from  the  center  is  9. 

97.  If  the  radius  of  a  sphere  is  20,  find : 
(a)  The  area  of  its  surface. 

(6)  The  area  of  a  zone  whose  altitude  is  2. 

(c)  The  edge  of  a  cube  inscribed  in  the  sphere. 

(d)  The  area  of  a  lune  whose  angle  is  80°. 

(e  )  The  area  of  a  spherical  triangle  whose  angles  are  75°,  53°,  72°. 

(/)  The  area  of  a  spherical  polygon  whose  angles  are  68°,  119°,  128°, 
147°,  150°. 

(gr)  The  area  of  a  birectangular  spherical  triangle  whose  vertex- 
angle  is  54°. 

(A)  The  area  of  a  zone  of  one  base  whose  altitude  is  5. 

(  i )  The  radius  of  a  sphere  whose  surface  is  four  times  as  large. 

(j)  The  volume  of  the  sphere. 

(fr)  The  volume  of  a  wedge  whose  angle  is  36°. 

(/)  The  volume  of  a  spherical  pyramid  whose  base  is  the  triangle 
of  exercise  (e). 

(w)  The  volume  of  the  spherical  sector  whose  base  is  the  zone  of 
exercise  (6). 

(n)  The  volume  of  the  spherical  cone  whose  base  is  the  zone  of 
exercise  (A). 

(o)  The  volume  of  a  spherical  segment  of  one  base,  whose  altitude 
is  6. 

(/>)  The  radius  of  a  sphere  whose  volume  is  four  times  as  large. 


INDEX   OF  DEFINITIONS 


(The  numbers  refer  to  pages.) 


Abbreviations,  16. 
Acute  angle,  13. 
Acute  triangle,  15. 
Adjacent  angles,  12. 
Alternate-exterior  angles,  38. 
Alternate-interior  angles,  38. 
Alternation,  141. 
Altitude,  of  cone,  345. 

of  cylinder,  337. 

of  frustum,  308,  346. 

of  parallelogram,  46. 

of  prism,  292. 

of  pyramid,  307. 

of  trapezoicl,  46. 

of  triangle,  33. 

of  zone,  386. 
Analysis,  127. 
Angle,  12. 

acute,  13. 

between  intersecting  curves,  361. 

birectangular  trihedral,  286. 

bisector  of,  14,  33. 

central,  80. 

central,  of  regular  polygon,  220. 

complement  of,  13. 

convex  polyhedral,  284. 

degree  of,  13. 

dihedral,  273. 

exterior,  of  polygon,  54. 

exterior,  of  triangle,  42. 

inscribed,  in  circle,  80. 

inscribed,  in  segment,  102. 

isosceles  trihedral,  286. 

measure  of,  102. 

oblique,  13. 

obtuse,  13. 

plane,  12. 

plane  angle  of  dihedral,  273. 

polyhedral,  284. 

rectangular  trihedral,  286. 

right,  12. 

sides  of,  12. 

spherical,  361. 


Angle,  supplement  of,  13. 

trihedral,  286. 

trirectangular  trihedral,  286. 

vertex  of,  12. 
Angles,  adjacent,  12. 

alternate-exterior,  38. 

alternate-interior,  38. 

complementary,  13. 

corresponding,  38. 

dihedral,  273. 

equal,  15. 

face,  of  polyhedral,  284. 

homologous,  of  polygons,  55. 

of  lune,  386. 

of  polygons,  54. 

of  spherical  triangle,  370. 

of  triangle,  14. 

polyhedral,  284. 

right  dihedral,  273. 

supplementary,  13. 

symmetrical  polyhedral,  285. 

vertical,  12. 
Antecedents,  140. 
Apothem,  220. 
Arc,  80. 

degree  of,  102. 

intercepted,  81. 

subtended,  81. 
Arcs,  similar,  228. 
Area,  187. 
Auxiliary  lines,  25. 
Axiom,  16. 

parallel,  36. 
Axioms,  16. 
Axis,  of  circle  of  sphere,  361. 

of  circular  cone,  345. 

of  symmetry,  58. 

Base,  of  cone,  345. 
of  figure,  45. 
of  pyramid,  307. 
of  spherical  pyramid,  387. 
of  spherical  sector,  387. 
405 


406 


INDEX   OF  DEFINITIONS 


Base,  of  triangle,  14. 
Bases,  of  cylinder,  337. 

of  parallelogram,  45. 

of  prism,  292. 

of  spherical  segment,  387. 

of  trapezoid,  45. 

of  zone,  386. 

Birectangular  spherical   triangle,   371. 
Birectangular  trihedral  angle,  286. 
Bisector  of  angle,  14,  33. 
Boundary,  12. 

Center,    figure   symmetrical    with    re- 
spect to,  58. 

of  circle,  79. 

of  circumference,  79. 

of  regular  polygon,  220. 

of  sphere,  360. 

of  symmetry,  58. 
Central  angle/ 80. 

of  regular  polygon,  220. 
Chord,  79. 
Circle,  79. 

angle  inscribed  in,  80. 

center  of,  79. 

circumscribed  about  polygon,  91. 

diameter  of,  79. 

inscribed  in  polygon,  92 

radius  of,  79. 

sector  of,  80. 

segment  of,  80. 

tangent  to,  79. 
Circles,  concentric,  80. 

equal,  80. 

escribed,  124. 

of  spheres,  361. 

tangent  externally,  80. 

tangent  internally,  80. 
Circular  cone,  345. 

axis  of,  345. 

right,  345. 
Circular  cylinder,  338. 

right,  338. 
Circumference,  79. 

center  of,  79. 
Circumscribed  circle,  91. 
Circumscribed  frustum  of  pyramid,  346. 
Circumscribed  polygon,  92. 
Circumscribed  prism,  312. 
Circumscribed  sphere,  361. 
Commensurable  quantities,  97. 
Common  tangent,  92. 
Complementary  angles,  13. 
Complement  of  angle,  13. 
Composition,  142. 


Composition  and  division,  143. 
Concave  polygon,  54. 
Concentric  circles,  80. 
Conclusion,  23. 
Concurrent  lines,  57. 
Cone,  345. 

altitude  of,  345. 

base  of,  345. 

circular,  345. 

circular,  axis  of,  345. 

frustum  of,  346. 

lateral  area  of,  345. 

lateral  area  of  frustum  of,  346. 

oblique  circular,  346. 

of  revolution,  346. 

of  revolution,  slant  height  of,  346. 

plane  tangent  to,  346. 

right  circular,  345. 

spherical,  387. 

total  area  of,  345. 

total  area  of  frustum  of,  346. 
Cones,  345. 

similar,  of  revolution,  346. 
Conical  surface,  345. 
Consequents,  140. 
Constant,  97. 
Construction,  115. 
Continued  proportion,  140. 
Converse  of  theorem,  25. 
Convex  polyhedral  angle,  284. 
Convex  polyhedron,  291. 
Corollary,  17. 
Corresponding  angles,  38. 
Cube,  293,  324. 
Curved  line,  79. 
Cylinder,  337. 

altitude  of,  337. 

bases  of,  337. 

circular,  338. 

lateral  area  of,  337. 

oblique,  338. 

of  revolution,  338. 

right,  338. 

right  circular,  338. 

right  section  of,  338. 

total  area  of,  337. 

Cylinders,   similar,  of  revolution,  338. 
Cylindrical  surface,  337. 

Decagon,  55. 
Degree,  of  angle,  13. 

of  arc,  102. 

spherical,  386. 
Demonstration,  17. 

elements  of,  23. 


INDEX  OF  DEFINITIONS 


407 


Determination  of  straight  line,  11. 

of  a  circle,  116. 
Determined,  plane,  252. 
Diagonal,  45. 

of  polyhedron,  291. 

of  spherical  polygon,  372. 
Diameter  of  circle,  79. 

of  sphere,  360. 
Dihedral  angle,  273. 

edge  of,  273. 

faces  of,  273. 

plane  angle  of,  273. 
Dihedral  angles,  273. 

right,  273. 

Dimensions  of  parallelepiped,  293. 
Direct  proportion,  160. 
Directrix,  337,  345. 
Discussion,  115. 

Distance    between    points    of    surface 
of  sphere,  361. 

between  two  points,  25. 

from  point,  to  a  line,  32. 

from  point  to  a  plane,  268. 
Division,  142. 
Dodecagon,  55. 
Dodecahedron,  291. 

regular,  324. 

Edge  of  dihedral  angle,  273. 
Edges  of  polyhedral  angle,  284. 

of  polyhedron,  291. 
Element  of  conical  surface,  345. 

of  cylindrical  surface,  337. 
Equal  angles,  15. 
Equal  circles,  83. 
Equal  dihedral  angles,  273. 
Equal  figures,  15. 
Equal  polygons,  55. 
Equal  polyhedral  angles,  285. 
Equal  solids,  293. 
Equal  spheres,  361. 
Equal  spherical  polygons,  372. 
Equiangular  polygon,  54. 
Equiangular  spherical  triangles,  371. 
Equiangular  triangle,  15. 
Equilateral  polygon,  54. 
Equilateral  spherical  triangles,  371. 
Equilateral  triangle,  15. 
Equivalent  figures,  187. 
Equivalent  solids,  293. 
Escribed  circles,  124. 
Exterior  angle,  of  polygon,  54. 

of  triangle,  42. 

Extreme  and  mean  ratio,  182. 
Extremes,  142. 


Face  angle  of  polyhedral  angle,  284. 
Faces,  of  dihedral  angle,  273. 

of  polyhedral  angle,  284. 

of  polyhedron,  291. 
Figure,  base  of,  45. 

rectilinear,  11. 

symmetrical,  58. 
Figures,  equal,  15. 

equivalent,  187. 

isoperimetric,  245. 
Foot  of  line,  251. 

of  perpendicular,  13. 
Fourth  proportional,  140. 
Frustum  of  cone,  346. 

altitude  of,  346. 

lateral  area  of,  346. 

midsection  of,  346. 

slant  height  of,  346. 

total  area  of,  346. 
Frustum  of  pyramid,  308. 

altitude  of,  308. 

circumscribed,  346. 

inscribed,  346. 

slant  height  of,  307. 

Generatrix,  337,  345. 
Geometrical  solid,  294. 
Geometry,   11. 

plane,  11.  - 

solid,  251. 
Great  circle  of  sphere,  361. 

axis  of,  361. 

Harmonic  division,  149. 
Hemisphere,  387. 
Heptagon,  55. 
Hexagon,  55. 
Hexahedron,  291. 

regular,  324. 

Homologous  angles,  in  polygons,  55. 
Homologous  parts,  15. 
Homologous  sides,  in  polygons,  55. 

in  triangles,  155. 
Hypotenuse,  15. 
Hypothesis,  23. 

[cosahedron,  291. 

regular,  324. 
Inclination,  268. 

[ncommensurable  quantities,  97. 
[ndirect  method,  36. 
Inscribed  angle,  in  circle,  80. 

in  segment,  102. 
Inscribed  circle,  92. 
Inscribed  frustum  of  pyramid,  346. 


408 


INDEX  OF   DEFINITIONS 


Inscribed  polygon,  91. 

Inscribed  prism,  312. 

Inscribed  prism  in  cylinder,  338. 

Inscribed  sphere  in  polyhedron,  361. 

Intercept,  to,  51,  81. 

Intercepted  arc,  81. 

Intersect,  to,  51. 

Intersection,  251. 

Inverse  proportion,  160. 

Inversion,  142. 

Isoperimetric  figures,  245. 

Isosceles  trapezoid,  45. 

Isosceles  triangle,  15. 

Isosceles  trihedral  angle,  286. 

Lateral  area,  of  cone,  345. 

of  cylinder,  337. 

of  frustum  of  cone,  346. 

of  prism,  292. 

of  pyramid,  307. 
Lateral  edges,  of  prism,  292. 

of  pyramid,  307. 
Lateral  faces,  of  prism,  292. 

of  pyramid,  307. 
Legs,  of  isosceles  trapezoid,  45. 

of  isosceles  triangle,  15. 

of  right  triangle,  15. 
Limit  of  variable,  97,  99. 
Limits,  theorem  of,  99. 
Line,  11. 

curved,  79. 

divided  harmonically,  149. 

divided    into    extreme    and    mean 
ratio,  182. 

foot  of,  251. 

inclination  of,  268. 

projection  of,  251. 

straight,  11. 

straight,  oblique  to  plane,  251. 

straight,  parallel  to  plane,  251. 

straight,     perpendicular    to     plane, 
251. 

tangent  to  circle,  79. 

tangent  to  circle  of  sphere,  361. 

tangent  to  sphere,  361. 
Lines,  auxiliary,  25. 

concurrent,  57. 

divided  proportionally,  147. 

oblique,  13. 

parallel,  36. 

perpendicular,  13. 
Locus,  60. 
Lune,  386. 

angles  of,  386. 

vertices  of,  386. 


Maximum,  245. 

Mean  proportional,  140. 

Means,  140. 

Measure,  of  angle,  102. 

of  quantity,  97. 

unit  of,  97. 
Median,  of  trapezoid,  46. 

of  triangle,  33. 
Method,  indirect,  36. 

of  exclusion,  36. 

Midsection  of  frustum  of  cone,  343. 
Minimum,  245. 
Motion,  12. 

Mutually  equiangular  polygons,  54. 
Mutually  equilateral  polygons,  54. 

Names  of  regular  polyhedrons,  324. 
N-gon,  55. 
Normal,  251. 

Oblique  angle,  13. 
Oblique  circular  cone,  346. 
Oblique  cylinder,  338. 
Oblique  lines,  13. 
Oblique  parallelepiped,  293. 
Oblique  prism,  292. 
Obtuse  angle,  13. 
Obtuse  triangle,  15. 
Octagon,  55. 
Octahedron,  291. 
regular,  324. 

Parallel  axiom,  36. 
Parallelepiped,  293. 

dimensions  of,  293. 

oblique,  293. 

rectangular,  293. 

right,  293. 
Parallel  lines,  36. 
Parallel  planes,  251. 
Parallelogram,  45. 

altitude  of,  46. 

bases  of,  45. 
Pentagon,  55 
Pentedecagon,  55. 
Perimeter,  92. 
Perpendicular,  13. 

foot  of,  13. 

Perpendicular  planes,  273. 
Pi  (*•),  227. 
Plane,  11,  251. 
Plane  angle,  13. 

Plane  angle  of  dihedral  angle,  273. 
Plane,  determined,  252. 

distance  from  point  to,  268. 


INDEX  OF   DEFINITIONS 


409 


Plane  Geometry,  11. 
Plane,  line  oblique  to,  251. 

line  parallel  to,  251. 

line  perpendicular  to,  251. 

tangent  to  cone,  346. 

tangent  to  cylinder,  338. 

tangent  to  sphere,  361. 
Plane  section,  284. 
Planes,  parallel,  251. 

perpendicular,  273. 
Point,  11. 

equally  distant  from  two  lines,  32. 

of  contact,  79,  361. 

of  tangency,  79,  361. 

projection  of,  163,  252. 
Points,  symmetrical,  58. 
Polar  distance,  361. 
Polar  triangle,  371. 
Poles  of  circle  of  sphere,  361. 
Polygon,  54. 

angles  of,  54. 

center  of  regular,  220. 

central  angle  of  regular,  220. 

circumscribed,  92. 

concave,  54. 

convex,  54. 

equiangular,  54. 

equilateral,  54. 

exterior  angle  of,  54. 

inscribed,  91. 

re-entrant,  54. 

regular,  217. 

vertices  of,  54. 
Polygons,  equal,  55. 

mutually  equiangular,  54. 

mutually  equilateral,  54. 

similar,  150. 
Polyhedral  angle,  284. 

convex,  284. 

edges  of,  284. 

face  angles  of,  284. 

faces  of,  284. 

plane  section  of,  284. 

vertex  of,  284. 
Polyhedral  angles,  284. 

equal,  285. 

symmetrical,  285. 

vertical,  285. 
Polyhedron,  291. 

convex,  291. 

diagonal  of,  291. 

edges  of,  291. 

faces  of,  291. 

regular,  323. 

sphere  inscribed  in,  361. 


Polyhedron,  vertices  of,  291. 
Polyhedrons,  similar,  323. 
Postulate,  17. 
Prism,  292. 

altitude  of,  292. 

bases  of,  292. 

circumscribed,  about  cylinder,  338. 

circumscribed,  about  pyramid,  312 

inscribed,  in  cylinder,  338. 

inscribed,  in  pyramid,  312. 

lateral  area  of,  292. 

lateral  edges  of,  292. 

lateral  faces  of,  292. 

oblique,  292. 

regular,  292. 

right,  292. 

right  section  of,  293. 

total  area  of,  292. 

triangular,  292. 

truncated,  292. 
Problem,  115. 
Projection,  of  line,  163,  252. 

of  point,  163,  252. 
Proof,  17. 
Proportion,  140. 

antecedents  of,  140. 

consequents  of,  140. 

continued,  140. 

direct,  160. 

extremes  of,  140. 

inverse,  160. 

means  of,  140. 

reciprocal,  160. 
Proportional,  fourth,  140. 

mean,  140. 

third,  140. 
Proposition,  115. 
Pyramid,  307. 

altitude  of,  307. 

altitude  of  frustum  of,  308. 

base  of,  307. 

circumscribed,  about  cone,  346. 

circumscribed,  frustum  of,  346. 

inscribed,  frustum  of,  346. 

inscribed,  in  cone,  346. 

frustum  of,  308. 

lateral  area  of,  307. 

lateral  edges  of,  307. 

lateral  faces  of,  307.  *. 

regular,  307. 

slant  height  of  frustum,  308. 

slant  height  of  regular,  307. 

spherical,  387. 

triangular,  307. 

total  area  of,  307. 


410 


INDEX  OF   DEFINITIONS 


Pyramid,  truncated,  308. 

vertex  of,  307. 
Pyramids,  307. 

Q.E.D.,  23. 
Q.E.F.,  116. 
Quadrant,  80,  361. 
Quadrilateral,  45. 

angles  of,  45. 

sides  of,  45. 

vertices  of,  45. 
Quantities,  commensurable,  97. 

incommensurable,  97. 

Radius,  of  circle,  79. 

of  regular  polygon,  220. 

of  sphere,  360. 
Ratio,  97,  140. 

extreme  and  mean,  182. 

series  of  equal,  140. 
Reciprocal  proportion,  160. 
Rectangle,  45. 

Rectangular  parallelepiped,  293. 
Rectangular  trihedral  angle,  286. 
Rectilinear  figure,  11. 
Reductio  ad  absurdum,  36. 
Re-entrant  polygon,  54. 
Regular  polygon,  217. 
Regular  polyhedrons,  323,  324. 
Regular  prism,  292. 
Regular  pyramid,  307. 

slant  height  of,  307. 
Revolution,  cone  of,  346. 

cylinder  of,  338. 

similar  cones  of,  346. 

similar  cylinders  of,  338. 
Rhomboid,  45. 
Rhombus,  45. 
Right  angle,  12. 
Right  circular  cone,  345. 
Right  circular  cylinder,  338. 
Right  cylinder,  338. 
Right  dihedral  angle,  273. 
Right  parallelepiped,  293. 
Right  prism,  292. 
Right  section,  of  cylinder,  338. 

of  prism,  293. 
Right  triangle,  15. 

Scalene  triangle,  15. 

Secant,  79. 

Sector,  of  circle,  80. 

spherical,  387. 
Sectors,  similar,  228. 
Segment,  base  of  spherical,  387. 


Segment,  of  circle,  80. 

spherical,  387. 
Segments,  of  line,  148. 

similar,  228. 
Semicircle,  80. 
Semicircumference,  80. 
Series  of  equal  ratios,  140. 
Sides,  homologous,  in  polygons,  55. 

homologous,  in  triangles,  155. 

of  angle,  12. 

of  polygon,  54. 

of  quadrilateral,  45. 

of  spherical  triangle,  370. 

of  triangle,  14. 
Similar  arcs,  228. 
Similar  cones  of  revolution,  346. 
Similar   cylinders    of   revolution,    338. 
Similar  polygons,  150. 
Similar  polyhedrons,  323. 
Similar  sectors,  228. 
Similar  segments,  228. 
Similar  triangles,  150. 
Slant  height,  of  cone,  346. 

of  frustum  of  cone,  346. 

of  frustum  of  pyramid,  308. 

of  regular  pyramid,  307. 
Small  circle  of  sphere,  361. 
Solid,  11,  251. 

geometrical,  294. 
Solid  geometry,  251. 

volume  of,  293. 
Solids,  equal,  293. 

equivalent,  293. 
Sphere,  360. 

axis  of  circle  of,  361. 

center  of,  360. 

circumscribed  about  polyhedron,  361. 

diameter  of,  360. 

great  circle  of,  361. 

inscribed  in  polyhedron,  361. 

line  tangent  to,  361. 

plane  tangent  to,  361. 

radius  of,  360. 

small  circle  of,  361. 
Spheres,  equal,  361. 

tangent,  361. 
Spherical  angle,  361. 
Spherical  cone,  387. 
Spherical  degree,  386. 
Spherical  excess,  387. 
Spherical  polygon,  372. 

diagonal  of,  372. 
Spherical  polygons,  equal,  372. 
Spherical  pyramid,  387. 

base  of,  387. 


INDEX  OF   DEFINITIONS 


411 


Spherical  pyramid,  vertex  of,  387. 
Spherical  sector,  387. 

base  of,  387. 
Spherical  segment,  387. 

altitude  of,  387. 

base  of,  387. 

of  one  base,  387. 
Spherical  surface,  360. 
Spherical  triangle,  370. 

angles  of,  370. 

birectangular,  371. 

sides  of,  370. 

symmetrical,  372. 

trirectangular,  371. 

unit  of  measure  of,  371. 

vertices  of,  370. 
Spherical  triangles,  370. 

mutually  equiangular,  371. 

mutually  equilateral,  371. 
Square,  45. 
Statement,  115. 
Straight  angle,  13. 
Straight  line,  11. 

divided,  extreme  and  mean  ratio,  182. 

divided  harmonically,  149. 

oblique  to  plane,  251. 

parallel  to  plane,  251. 

perpendicular  to  plane,  251. 

tangent  to  circle,  79. 

tangent  to  sphere,  361. 
Straight  lines,  divided  proportionally, 

147. 

Subtend,  to,  80. 
Subtended  arc,  81. 
Superposition,  15. 
Supplement  of  angle,  13. 
Supplementary  angles,  13. 
Surface,  11. 

conical,  345. 

cylindrical,  337. 

spherical,  360. 
Surfaces,  251. 
Symbols,  16. 

Symmetrical  polyhedral  angles,  285. 
Symmetrical  spherical  triangles,  372. 
Symmetry,  58. 

axis  of,  58. 

center  of,  58. 

Tangent,  79. 

circles,  80. 

spheres,  361. 
Tetrahedron,  291. 
Theorem,  17. 

converse  of,  25. 


Theorem,  elements  of,  23. 

of  limits,  99. 
Third  proportional,  140. 
Total  area,  of  cone,  345. 

of  cylinder,  337. 

of  frustum  of  cone,  346. 

of  prism,  292. 

of  pyramid,  307. 
Transversal,  38. 
Trapezium,  45. 
Trapezoid,  45. 

altitude  of,  46. 

bases  of,  45. 

isosceles,  45. 

legs  of,  45. 

median  of,  46. 
Triangle,  14. 

acute,  15. 

altitude  of,  33. 

angles  of,  14. 

base  of,  14. 

equiangular,  15. 

equilateral,  15. 

exterior  angle  of,  42. 

isosceles,  15. 

median  of,  33 

obtuse,  15. 

right,  15. 

scalene,  15. 

sides  of,  14. 

spherical,  370. 

vertex  of,  14. 

vertices  of,  14. 
Triangles,  similar,  150. 
Triangular  prism,  292. 
Triangular  pyramid,  307. 
Trihedral  angle,  286. 

birectangular,  286. 

isosceles,  286. 

trirectangular,  286. 

Trirectangular  spherical  triangle,  371, 
Trirectangular  trihedral  angle,  286. 
Truncated  prism,  292. 
Truncated  pyramid,  308. 

Unit,  of  measure,  97. 

of  spherical  triangle,  371. 
of  surface,  187. 
of  volume,  293. 

Variable,  97. 

limit  of,  97,  99. 
Vertex  angle,  14. 
Vertex,  of  angle,  12. 

of  polyhedral  angle,  284. 


412 


INDEX  OF   DEFINITIONS 


Vertex,  of  pyramid,  307. 

of  spherical  pyramid,  387. 

of  triangle,  14. 
Vertical  angles,  12. 
Vertical  dihedral  angles,  273. 
Vertical  polyhedral  angles,  285. 
Vertices,  of  June,  386. 

of  polygon,  54. 

of  polyhedron,  291. 

of  quadrilateral,  45. 

of  spherical  triangle,  370. 


Vertices,  of  triangle,  14. 
Volume,  of  solid,  293. 
unit  of,  293. 

Wedge,  spherical,  387. 

Zone,  386. 

altitude  of,  386. 
bases  of,  386. 
of  one  base,  386. 





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